Drawing a binary tree

Hi guys ;
I'm trying to write a Java GUI that draws a binary tree on a JPanel.
the tree is dynamically filled while user is inserting nodes.
I need a quit robust Implementation so that the tree is properly drawn with lines connecting nodes and most IMPORTANT ; the panel can expend to fit the tree (dynamically).
can any one tell me how to get started ?
Many thanks !

My advice is to use a graphics framework, e.g. look at Piccolo:
http://www.cs.umd.edu/hcil/piccolo/index.shtml
If you put your panel in a piccolo scrollpane it will be able to expand in all directions, something that is very hard to achieve otherwise.
OTOH Piccolo is not a dedicated graph framework (which knows about the distinction of nodes and edges and how to connect them). It comes with an example, though how to implement that.
There are more dedicated graph frameworks, e.g. JGraph:
http://sourceforge.net/projects/jgraph/

Similar Messages

Binary tree using inorder traversal algorithm

I need to draw a Binary tree using Inorder traversal algorithm.. in java applets using rectangles and ellipse objects.. can anyone help me about this ? I will be really thankful to yu

I think this is my psychic Sunday, so here's my attempt at answering
this vaguely stated question. First have a look at this fine piece of
ASCII art:--------------------------------------------   ^         ^
| A                    T                     |   |         |
|                                            |   |        root
---------L-----------------------R---------+   |         v
| B                    | C                  |   |
|                      |                     |   |
|                      |                     | height
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
-------------------------------------------+   v
<--------------------width------------------->Suppose you have a rectangle 'width' wide and 'height' high. The root
of the tree can be drawn at position T and (if present), two lines can be
drawn to the positions L and R respectively. Now recursively call your
same method using the computed rectangle A and B for the left and
right subtrees L and R respectively.
kind regards,
Jos

Constructing Binary tree

So this is my first post here and i am beginning to like this forum.
Can anybody help me on how to construct Binary tree from inorder and postorder traversals,i just want to know the algorithm so that i can apply it.
Please help.

I would like to pick a minor little nit with this analysis. The algorithm that has been proposed assumes that all the nodes are all distinct and that having selected the root from the end of the post-order listing that it is POSSIBLE to find it in the in-order list. What if you find multiple copies of this node?
If multiple copies of the root are found, you must have a method to distinguish, which one is the proper dividing point. In the worst possible case, the problem can not be solved at all. For example suppose my post-order and my in-order lists were these:
a a a a a
a a a a a
The shape of the tree is indeterminant in this case.
If you allow different tree nodes to contain identical values your recursive algorithm needs some modification.
The fix is this:
1) allow your recursive algorithm to fail (and report back any success or failure)
This can and happen if the two lists that you passed in are incompatible. For example they could have different nodes in them.
2) when you pick the root off the end of the post order list, you search for it in the in-order list, you could find multiple matches or you could find no matches. You must explore each of these independently because each one could lead to a possible different solution, or could lead to no solution. Of course in the case of no matches, you must report back a failure.
Depending on your needs, you can either stop the first time that you have successfully assembled a tree that matches the two supplied inputs, or you can have it grind on and have it enumerate all the possible tree arrangements that could have generated from the two traversals that you started with.
It might help to visualize if you write out all the possible trees with just the three nodes AAB. There are 15 of them, 5 with B at the root, 5 with A at the root and B in the left and 5 with B in the right. It is easy to draw the trees and to immediately write both their in-order and their post-order traversals.
Any traversal is just a list of the 3 nodes and there are 3 arrangements, AAB, ABA, and BAA. There are exactly 9 ordered pairs of these traversals so you can't get all 15 trees from the 9 pairs.
Sho nuff, three ordered pairs are unambiguous and generate a single unique tree(e.g. in:BAA post:ABA) and six of them come from ambiguous pairs of trees (e.g. in:ABA post:ABA - you can't tell if this is a tree ABA all leaning to the left or all leaning to the right)
Enjoy

How to crossover this binary tree..?

You can view detail http://www.codeguru.com/forum/showthread.php?s=bb4cf7ad2b18a5115e8bd6ab3a4e9d17&t=470868
[nha khoa|http://www.sieuthi77.com/main/nhakhoa.html] .com/forum/showthread.php?s=bb4cf7ad2b18a5115e8bd6ab3a4e9d17&t=470868
I have these classes which model a tree (binary). the thing is i cant figure out how i can set the elements of individual nodes in that tree. i can get individual elements but i cannot set them due to the strucutre of the tree and how it is implemented. The changes that I make to these classes should be as less as possible, because i have an algorithm which generates the tree structure randomly, plus i can evaluate the tree easily by using recursion. The only left thing to do is CROSSOVER but how?!?
here are the classes which model my binary tree:
Code:
public abstract class Node implements Cloneable{
abstract double evaluate(VariableInput v);
abstract String print();
abstract int getNumberOfNodes();
abstract ArrayList<Object> getChildren();
@Override
public Node clone(){
Node copy;
try {
copy = (Node) super.clone();
} catch (CloneNotSupportedException unexpected) {
throw new AssertionError(unexpected);
//In an actual implementation of this pattern you might now change references to
//the expensive to produce parts from the copies that are held inside the prototype.
return copy;
Code:
public class UnaryNode extends Node {
private UnaryFunction operator;
private Node left;
public UnaryNode(UnaryFunction op, Node terminal) {
operator = op;
this.left = terminal;
public String print(){
String r = "(" + operator.toString()+ " " + left.print() + ")";
return r;
void setLeft(Node left) {
this.left = left;
@Override
int getNumberOfNodes() {
return 1 + left.getNumberOfNodes();
Node getLeft() {
return left;
ArrayList<Object> getChildren() {
ArrayList<Object> arr = new ArrayList<Object>();
arr.add(this);
arr.addAll(left.getChildren());
return arr;
Code:
public class BinaryNode extends Node {
private BinaryFunction operator;
private Node left;
private Node right;
public BinaryNode(BinaryFunction op, Node left, Node right) {
operator = op;
this.left = left;
this.right = right;
public String print(){
String r = "(" + operator.toString()+ " " + left.print() + " " + right.print()+")";
return r;
public void setLeft(Node left){
this.left = left;
public void setRight(Node right){
this.right = right;
@Override
int getNumberOfNodes() {
return 1 + left.getNumberOfNodes() + right.getNumberOfNodes();
Node getRight() {
return right;
Node getLeft() {
return left;
@Override
ArrayList<Object> getChildren() {
ArrayList<Object> arr = new ArrayList<Object>();
arr.add(this);
arr.addAll(left.getChildren());
arr.addAll(right.getChildren());
return arr;
public class NumericNode extends Node{
private double value;
public NumericNode(double v){
value = v;
@Override
double evaluate(VariableInput c) {
return value;
public String print(){
String r = "" + value;
return r;
@Override
int getNumberOfNodes() {
return 1;
@Override
ArrayList<Object> getChildren() {
ArrayList<Object> arr = new ArrayList<Object>();
arr.add(new NumericNode(value));
return arr;
}p.s. I have this get children method which return a list of REFERENCES to all the nodes in the tree, but if i change any of them it wont have an effect to the tree itself because they are references.
Any ideas or codes will be much appreciated. Thanks!

What? Changes to what a node is referencing will be reflected in the tree, unless your getChildren is returning a copy, like in NumericNode.
Kaj

Need Help with a String Binary Tree

Hi, I need the code to build a binary tree with string values as the nodes....i also need the code to insert, find, delete, print the nodes in the binarry tree
plssss... someone pls help me on this
here is my code now:
// TreeApp.java
// demonstrates binary tree
// to run this program: C>java TreeApp
import java.io.*; // for I/O
import java.util.*; // for Stack class
import java.lang.Integer; // for parseInt()
class Node
     //public int iData; // data item (key)
     public String iData;
     public double dData; // data item
     public Node leftChild; // this node's left child
     public Node rightChild; // this node's right child
     public void displayNode() // display ourself
          System.out.print('{');
          System.out.print(iData);
          System.out.print(", ");
          System.out.print(dData);
          System.out.print("} ");
} // end class Node
class Tree
     private Node root; // first node of tree
     public Tree() // constructor
     { root = null; } // no nodes in tree yet
     public Node find(int key) // find node with given key
     {                           // (assumes non-empty tree)
          Node current = root; // start at root
          while(current.iData != key) // while no match,
               if(key < current.iData) // go left?
                    current = current.leftChild;
               else // or go right?
                    current = current.rightChild;
               if(current == null) // if no child,
                    return null; // didn't find it
          return current; // found it
     } // end find()
     public Node recfind(int key, Node cur)
          if (cur == null) return null;
          else if (key < cur.iData) return(recfind(key, cur.leftChild));
          else if (key > cur.iData) return (recfind(key, cur.rightChild));
          else return(cur);
     public Node find2(int key)
          return recfind(key, root);
public void insert(int id, double dd)
Node newNode = new Node(); // make new node
newNode.iData = id; // insert data
newNode.dData = dd;
if(root==null) // no node in root
root = newNode;
else // root occupied
Node current = root; // start at root
Node parent;
while(true) // (exits internally)
parent = current;
if(id < current.iData) // go left?
current = current.leftChild;
if(current == null) // if end of the line,
{                 // insert on left
parent.leftChild = newNode;
return;
} // end if go left
else // or go right?
current = current.rightChild;
if(current == null) // if end of the line
{                 // insert on right
parent.rightChild = newNode;
return;
} // end else go right
} // end while
} // end else not root
} // end insert()
public void insert(String id, double dd)
     Node newNode = new Node(); // make new node
     newNode.iData = id; // insert data
     newNode.dData = dd;
     if(root==null) // no node in root
          root = newNode;
     else // root occupied
          Node current = root; // start at root
          Node parent;
          while(true) // (exits internally)
               parent = current;
               //if(id < current.iData) // go left?
               if(id.compareTo(current.iData)>0)
                    current = current.leftChild;
                    if(current == null) // if end of the line,
                    {                 // insert on left
                         parent.leftChild = newNode;
                         return;
               } // end if go left
               else // or go right?
                    current = current.rightChild;
                    if(current == null) // if end of the line
                    {                 // insert on right
                         parent.rightChild = newNode;
                         return;
               } // end else go right
          } // end while
     } // end else not root
} // end insert()
     public Node betterinsert(int id, double dd)
          // No duplicates allowed
          Node return_val = null;
          if(root==null) {       // no node in root
               Node newNode = new Node(); // make new node
               newNode.iData = id; // insert data
               newNode.dData = dd;
               root = newNode;
               return_val = root;
          else // root occupied
               Node current = root; // start at root
               Node parent;
               while(current != null)
                    parent = current;
                    if(id < current.iData) // go left?
                         current = current.leftChild;
                         if(current == null) // if end of the line,
                         {                 // insert on left
                              Node newNode = new Node(); // make new node
                              newNode.iData = id; // insert data
                              newNode.dData = dd;
                              return_val = newNode;
                              parent.leftChild = newNode;
                    } // end if go left
                    else if (id > current.iData) // or go right?
                         current = current.rightChild;
                         if(current == null) // if end of the line
                         {                 // insert on right
                              Node newNode = new Node(); // make new node
                              newNode.iData = id; // insert data
                              newNode.dData = dd;
                              return_val = newNode;
                              parent.rightChild = newNode;
                    } // end else go right
                    else current = null; // duplicate found
               } // end while
          } // end else not root
          return return_val;
     } // end insert()
     public boolean delete(int key) // delete node with given key
          if (root == null) return false;
          Node current = root;
          Node parent = root;
          boolean isLeftChild = true;
          while(current.iData != key) // search for node
               parent = current;
               if(key < current.iData) // go left?
                    isLeftChild = true;
                    current = current.leftChild;
               else // or go right?
                    isLeftChild = false;
                    current = current.rightChild;
               if(current == null)
                    return false; // didn't find it
          } // end while
          // found node to delete
          // if no children, simply delete it
          if(current.leftChild==null &&
               current.rightChild==null)
               if(current == root) // if root,
                    root = null; // tree is empty
               else if(isLeftChild)
                    parent.leftChild = null; // disconnect
               else // from parent
                    parent.rightChild = null;
          // if no right child, replace with left subtree
          else if(current.rightChild==null)
               if(current == root)
                    root = current.leftChild;
               else if(isLeftChild)
                    parent.leftChild = current.leftChild;
               else
                    parent.rightChild = current.leftChild;
          // if no left child, replace with right subtree
          else if(current.leftChild==null)
               if(current == root)
                    root = current.rightChild;
               else if(isLeftChild)
                    parent.leftChild = current.rightChild;
               else
                    parent.rightChild = current.rightChild;
               else // two children, so replace with inorder successor
                    // get successor of node to delete (current)
                    Node successor = getSuccessor(current);
                    // connect parent of current to successor instead
                    if(current == root)
                         root = successor;
                    else if(isLeftChild)
                         parent.leftChild = successor;
                    else
                         parent.rightChild = successor;
                    // connect successor to current's left child
                    successor.leftChild = current.leftChild;
                    // successor.rightChild = current.rightChild; done in getSucessor
               } // end else two children
          return true;
     } // end delete()
     // returns node with next-highest value after delNode
     // goes to right child, then right child's left descendents
     private Node getSuccessor(Node delNode)
          Node successorParent = delNode;
          Node successor = delNode;
          Node current = delNode.rightChild; // go to right child
          while(current != null) // until no more
          {                                 // left children,
               successorParent = successor;
               successor = current;
               current = current.leftChild; // go to left child
          // if successor not
          if(successor != delNode.rightChild) // right child,
          {                                 // make connections
               successorParent.leftChild = successor.rightChild;
               successor.rightChild = delNode.rightChild;
          return successor;
     public void traverse(int traverseType)
          switch(traverseType)
          case 1: System.out.print("\nPreorder traversal: ");
               preOrder(root);
               break;
          case 2: System.out.print("\nInorder traversal: ");
               inOrder(root);
               break;
          case 3: System.out.print("\nPostorder traversal: ");
               postOrder(root);
               break;
          System.out.println();
     private void preOrder(Node localRoot)
          if(localRoot != null)
               localRoot.displayNode();
               preOrder(localRoot.leftChild);
               preOrder(localRoot.rightChild);
     private void inOrder(Node localRoot)
          if(localRoot != null)
               inOrder(localRoot.leftChild);
               localRoot.displayNode();
               inOrder(localRoot.rightChild);
     private void postOrder(Node localRoot)
          if(localRoot != null)
               postOrder(localRoot.leftChild);
               postOrder(localRoot.rightChild);
               localRoot.displayNode();
     public void displayTree()
          Stack globalStack = new Stack();
          globalStack.push(root);
          int nBlanks = 32;
          boolean isRowEmpty = false;
          System.out.println(
          while(isRowEmpty==false)
               Stack localStack = new Stack();
               isRowEmpty = true;
               for(int j=0; j<nBlanks; j++)
                    System.out.print(' ');
               while(globalStack.isEmpty()==false)
                    Node temp = (Node)globalStack.pop();
                    if(temp != null)
                         System.out.print(temp.iData);
                         localStack.push(temp.leftChild);
                         localStack.push(temp.rightChild);
                         if(temp.leftChild != null ||
                              temp.rightChild != null)
                              isRowEmpty = false;
                    else
                         System.out.print("--");
                         localStack.push(null);
                         localStack.push(null);
                    for(int j=0; j<nBlanks*2-2; j++)
                         System.out.print(' ');
               } // end while globalStack not empty
               System.out.println();
               nBlanks /= 2;
               while(localStack.isEmpty()==false)
                    globalStack.push( localStack.pop() );
          } // end while isRowEmpty is false
          System.out.println(
     } // end displayTree()
} // end class Tree
class TreeApp
     public static void main(String[] args) throws IOException
          int value;
          double val1;
          String Line,Term;
          BufferedReader input;
          input = new BufferedReader (new FileReader ("one.txt"));
          Tree theTree = new Tree();
     val1=0.1;
     while ((Line = input.readLine()) != null)
          Term=Line;
          //val1=Integer.parseInt{Term};
          val1=val1+1;
          //theTree.insert(Line, val1+0.1);
          val1++;
          System.out.println(Line);
          System.out.println(val1);
theTree.insert(50, 1.5);
theTree.insert(25, 1.2);
theTree.insert(75, 1.7);
theTree.insert(12, 1.5);
theTree.insert(37, 1.2);
theTree.insert(43, 1.7);
theTree.insert(30, 1.5);
theTree.insert(33, 1.2);
theTree.insert(87, 1.7);
theTree.insert(93, 1.5);
theTree.insert(97, 1.5);
          theTree.insert(50, 1.5);
          theTree.insert(25, 1.2);
          theTree.insert(75, 1.7);
          theTree.insert(12, 1.5);
          theTree.insert(37, 1.2);
          theTree.insert(43, 1.7);
          theTree.insert(30, 1.5);
          theTree.insert(33, 1.2);
          theTree.insert(87, 1.7);
          theTree.insert(93, 1.5);
          theTree.insert(97, 1.5);
          while(true)
               putText("Enter first letter of ");
               putText("show, insert, find, delete, or traverse: ");
               int choice = getChar();
               switch(choice)
               case 's':
                    theTree.displayTree();
                    break;
               case 'i':
                    putText("Enter value to insert: ");
                    value = getInt();
                    theTree.insert(value, value + 0.9);
                    break;
               case 'f':
                    putText("Enter value to find: ");
                    value = getInt();
                    Node found = theTree.find(value);
                    if(found != null)
                         putText("Found: ");
                         found.displayNode();
                         putText("\n");
                    else
                         putText("Could not find " + value + '\n');
                    break;
               case 'd':
                    putText("Enter value to delete: ");
                    value = getInt();
                    boolean didDelete = theTree.delete(value);
                    if(didDelete)
                         putText("Deleted " + value + '\n');
                    else
                         putText("Could not delete " + value + '\n');
                    break;
               case 't':
                    putText("Enter type 1, 2 or 3: ");
                    value = getInt();
                    theTree.traverse(value);
                    break;
               default:
                    putText("Invalid entry\n");
               } // end switch
          } // end while
     } // end main()
     public static void putText(String s)
          System.out.print(s);
          System.out.flush();
     public static String getString() throws IOException
          InputStreamReader isr = new InputStreamReader(System.in);
          BufferedReader br = new BufferedReader(isr);
          String s = br.readLine();
          return s;
     public static char getChar() throws IOException
          String s = getString();
          return s.charAt(0);
     public static int getInt() throws IOException
          String s = getString();
          return Integer.parseInt(s);
} // end class TreeApp

String str = "Hello";
          int index = 0, len = 0;
          len = str.length();
          while(index < len) {
               System.out.println(str.charAt(index));
               index++;
          }

A Binary Tree Implementation in ABAP

Hi,
Can any one explaine me how to create a binary tree of random numbers with dynamic objects.
Thanks,
Manjula.

Hi manjula,
This sample code uses dynamic objects to create a binary tree of random numbers as per your requirement ...pls go through It.
It stores numbers on the left node or right node depending on the value comparison with the current value. There are two recursive subrotines used for the building of the tree and printing through the tree.
For comparison purpose, the same random numbers are stored and sorted in an internal table and printed.
*& Report YBINTREE - Build/Print Binary Tree of numbers *
report ybintree .
types: begin of stree,
value type i,
left type ref to data,
right type ref to data,
end of stree.
data: tree type stree.
data: int type i.
data: begin of rnd occurs 0,
num type i,
end of rnd.
start-of-selection.
do 100 times.
generate random number between 0 and 100
call function 'RANDOM_I4'
exporting
rnd_min = 0
rnd_max = 100
importing
rnd_value = int.
store numbers
rnd-num = int.
append rnd.
build binary tree of random numbers
perform add_value using tree int.
enddo.
stored numbers are sorted for comparison
sort rnd by num.
print sorted random numbers
write: / 'Sorted Numbers'.
write: / '=============='.
skip.
loop at rnd.
write: rnd-num.
endloop.
skip.
print binary tree. This should give the same result
as the one listed from the internal table
write: / 'Binary Tree List'.
write: / '================'.
skip.
perform print_value using tree.
skip.
*& Form add_value
text - Build tree with value provided
-->TREE text
-->VAL text
form add_value using tree type stree val type i.
field-symbols: <ltree> type any.
data: work type stree.
if tree is initial. "When node has no values
tree-value = val. " assign value
clear: tree-left, tree-right.
create data tree-left type stree. "Create an empty node for left
create data tree-right type stree. "create an empty node for right
else.
if val le tree-value. "if number is less than or equal
assign tree-left->* to <ltree>. "assign the left node to fs
call add_value recursively with left node
perform add_value using <ltree> val.
else. "if number is greater
assign tree-right->* to <ltree>. "assign the right node to fs
call add_value recursively with right node
perform add_value using <ltree> val.
endif.
endif.
endform. "add_value
*& Form print_value
text - traverse tree from left-mid-right order
automatically this will be sorted list
-->TREE text
form print_value using tree type stree.
field-symbols: <ltree> type any.
if tree is initial. "node is empty
else. "non-empty node
assign tree-left->* to <ltree>. "left node
perform print_value using <ltree>. "print left
write: tree-value. "print the current value
assign tree-right->* to <ltree>. "right node
perform print_value using <ltree>. "print right
endif.
endform. "print_value
pls reward if helps,
regards.

Having trouble finding the height of a Binary Tree

Hi, I have an ADT class called DigitalTree that uses Nodes to form a binary tree; each subtree only has two children at most. Each node has a "key" that is just a long value and is placed in the correct position on the tree determined by its binary values. For the height, I'm having trouble getting an accurate height. With the data I'm using, I should get a height of 5 (I use an array of 9 values/nodes, in a form that creates a longest path of 5. The data I use is int[] ar = {75, 37, 13, 70, 75, 90, 15, 13, 2, 58, 24} ). Here is my code for the whole tree. If someone could provide some tips or clues to help me obtain the right height value, or if you see anything wrong with my code, it would be greatly aprpeciated. Thanks!
public class DigitalTree<E> implements Copyable
   private Node root;
   private int size;
   public DigitalTree()
       root = null;
       size = 0;
   public boolean add(long k)
       if(!contains(k))
            if(this.size == 0)
                root = new Node(k);
                size++;
                System.out.println(size + " " + k);
            else
                String bits = Long.toBinaryString(k);
                //System.out.println(bits);
                return add(k, bits, bits.length(), root);
            return true;
       else
           return false;
   private boolean add(long k, String bits, int index, Node parent)
       int lsb;
       try
           lsb = Integer.parseInt(bits.substring(index, index - 1));
       catch(StringIndexOutOfBoundsException e)
           lsb = 0;
       if(lsb == 0)
           if(parent.left == null)
               parent.left = new Node(k);
               size++;
               //System.out.println(size + " " + k);
               return true;
           else
               return add(k, bits, index-1, parent.left);
       else
           if(parent.right == null)
               parent.right = new Node(k);
               size++;
               //System.out.println(size + " " + k);
               return true;
           else
               return add(k, bits, index-1, parent.right);
   public int height()
       int leftHeight = 0, rightHeight = 0;
       return getHeight(root, leftHeight, rightHeight);
   private int getHeight(Node currentNode, int leftHeight, int rightHeight)
       if(currentNode == null)
           return 0;
       //else
       //    return 1 + Math.max(getHeight(currentNode.right), getHeight(currentNode.left));
       if(currentNode.left == null)
           leftHeight = 0;
       else
           leftHeight = getHeight(currentNode.left, leftHeight, rightHeight);
       if(currentNode.right == null)
           return 1 + leftHeight;
       return 1 + Math.max(leftHeight, getHeight(currentNode.right, leftHeight, rightHeight));
   public int size()
       return size;
   public boolean contains(long k)
        String bits = Long.toBinaryString(k);
        return contains(k, root, bits, bits.length());
   private boolean contains(long k, Node currentNode, String bits, int index)
       int lsb;
       try
           lsb = Integer.parseInt(bits.substring(index, index - 1));
       catch(StringIndexOutOfBoundsException e)
           lsb = 0;
       if(currentNode == null)
           return false;
       else if(currentNode.key == k)
           return true;
       else
           if(lsb == 0)
               return contains(k, currentNode.left, bits, index-1);
           else
               return contains(k, currentNode.right, bits, index-1);
   public Node locate(long k)
        if(contains(k))
            String bits = Long.toBinaryString(k);
            return locate(k, root, bits, bits.length());
        else
            return null;
   private Node locate(long k, Node currentNode, String bits, int index)
       int lsb;
       try
           lsb = Integer.parseInt(bits.substring(index, index - 1));
       catch(StringIndexOutOfBoundsException e)
           lsb = 0;
       if(currentNode.key == k)
           return currentNode;
       else
           if(lsb == 0)
               return locate(k, currentNode.left, bits, index-1);
           else
               return locate(k, currentNode.right, bits, index-1);
   public Object clone()
       DigitalTree<E> treeClone = null;
       try
           treeClone = (DigitalTree<E>)super.clone();
       catch(CloneNotSupportedException e)
           throw new Error(e.toString());
       cloneNodes(treeClone, root, treeClone.root);
       return treeClone;
   private void cloneNodes(DigitalTree treeClone, Node currentNode, Node cloneNode)
       if(treeClone.size == 0)
           cloneNode = null;
           cloneNodes(treeClone, currentNode.left, cloneNode.left);
           cloneNodes(treeClone, currentNode.right, cloneNode.right);
       else if(currentNode != null)
           cloneNode = currentNode;
           cloneNodes(treeClone, currentNode.left, cloneNode.left);
           cloneNodes(treeClone, currentNode.right, cloneNode.right);
   public void printTree()
       System.out.println("Tree");
   private class Node<E>
      private long key;
      private E data;
      private Node left;
      private Node right;
      public Node(long k)
         key = k;
         data = null;
         left = null;
         right = null;
      public Node(long k, E d)
         key = k;
         data = d;
         left = null;
         right = null;
      public String toString()
         return "" + key;
}

You were on the right track with the part you commented out; first define a few things:
1) the height of an empty tree is nul (0);
2) the height of a tree is one more than the maximum of the heights of the left and right sub-trees.
This translates to Java as a recursive function like this:
int getHeight(Node node) {
   if (node == null) // definition #1
      return 0;
   else // definition #2
      return 1+Math.max(getHeight(node.left), getHeight(node.right));
}kind regards,
Jos

Help with a binary tree

I'm writing a binary tree class and am having some trouble with the Insert function. Here is the code for the TreeNode class...
public class TreeNode
     TreeNode Left;
     TreeNode Right;
     String Name;
     public TreeNode(String NodeName)
          Left = null;
          Right = null;
          Name = NodeName;
}And this is the code for the Tree class...
public class Tree
     TreeNode Root;
     public Tree(String RootNode)
          Root = new TreeNode(RootNode);
     public void Insert(String Name)
          InsertNode(Root, Name);
     public void InsertNode(TreeNode t, String NodeName)
          if (t == null)
               t = new TreeNode(NodeName);
          else
               if (NodeName.compareTo(t.Name) < 0)
                    InsertNode(t.Left, NodeName);
               else if (NodeName.compareTo(t.Name) > 0)
                    InsertNode(t.Right, NodeName);
               else if (NodeName.compareTo(t.Name) == 0)
                    System.out.println("Entered node that was already in Tree");
}When I enter a new node into a Tree containing just the root, it follows the recursion through once, then creates the new TreeNode as it should. However, the new node is not really recognized by the tree because when I try to insert another node, it only finds the root in the tree and only goes through one recursion. What's wrong?

I believe t.Left (or t.Right) is getting set in the line
t = new TreeNode(NodeName);
Since it is a recursive function, when it is called the second time, the "t" that is passed in is actually the original t.left (I think), which is getting set then.

Searching for a certain binary tree from another tree........

I have been struggling for a tree search problem for a good while. Now I decide to ask you experts for a better solution :-).
Given a binary tree A. We know that every Node of A has two pointers. Leaves of A can be tested by if(node.right = =node). Namely, The right pointer of every leaf node points to itself. (The left pointer points to the node sits on the left side of the leaf in the same depth. and the leafmost node points to the root. I do no think this information is important, am i right?).
Tree B has a similar structure.
The node used for both A and B.
Node{
Node left;
Node right;
My question is how to test if tree B is a subtree of A and if it is, returns the node in A that corresponds to the root of B. otherwise, return null.
So, the method should look like:
public Node search(Node rootOfA, Node rootOfB){
I know a simple recursive fuction can do the job. The question is all about the effciency....
I am wonderring if this is some kind of well-researched problem and if there has been a classical solution.
Anyone knows of that? Any friend can give a sound solution?
Thank you all in advance.
Jason
Message was edited by:
since81

I'm not too sure if this would help but there goes.
I think a recursive function will be the easiest to implement (but not the most efficient). In terms of recursive function if you really want to add performance. You could implement your own stack and replace the recursive function with the use of this stack (since really the benefit of recursive function is that it manages its own stack). A non-recursive function with customized well implemented stack will be much more efficient but your code will become more ugly too (due to so many things to keep track of).
Is tree B a separate instance of the binary tree? If yes then how can Tree B be a subset/subtree of tree A (since they are two separate "trees" or instances of the binary tree). If you wish to compare the data /object reference of Tree B's root node to that of Tree A's then the above method would be the most efficient according to my knowledge. You might have to use a Queue but I doubt it. Stack should be able to replace your recursive function to a better more efficient subroutine but you will have to manage using your own stack (as mentioned above). Your stack will behave similar to the recursive stack to keep track of the child/descendant/parent/root node and any other references that you may use otherwise.
:)

Binary Tree Help

I have a project where I need to build a binary tree with random floats and count the comparisons made. The problem I'm having is I'm not sure where to place the comaprison count in my code. Here's where I have it:
public void insert(float idata)
          Node newNode = new Node();
          newNode.data = idata;
          if(root==null)
               root = newNode;
          else
               Node current = root;
               Node parent;
               while(true)
                    parent = current;
                    if(idata < current.data)
                         comp++;
                         current = current.leftc;
                         if(current == null)
                              parent.leftc = newNode;
                              return;
                    else
                         current = current.rightc;
                         if(current == null)
                              parent.rightc = newNode;
                              return;
     }//end insertDo I have it in the right place? Also, if I'm building the tree for 10,000 numbers would I get a new count for each level or would I get one count for comparisons?? I'd appreciate anyone's help on this.

You never reset the comp variable, so each timeyou
insert into the tree, it adds the number ofinserts
to the previous value.Yes, or something like that. I'm not sure what theOP
really means.Yeah, it's hard to be sure without seeing the rest of
the code.Sorry, I thought I had already posted my code for you to look at.
Here's a copy of it:
class Node
     public float data;
     public Node leftc;
     public Node rightc;
public class Btree
     private Node root;
     private int comp;
     public Btree(int value)
          root = null;
     public void insert(float idata)
          Node newNode = new Node();
          newNode.data = idata;
          if(root==null)
               root = newNode;
          else
               Node current = root;
               Node parent;
               while(true)
                    parent = current;
                    comp++;
                    if(idata < current.data)
                         current = current.leftc;
                         if(current == null)
                              parent.leftc = newNode;
                              return;
                    else
                         current = current.rightc;
                         if(current == null)
                              parent.rightc = newNode;
                              return;
     }//end insert
    public void display()
         //System.out.print();
         System.out.println("");
         System.out.println(comp);
} //end Btree
class BtreeApp
     public static void main(String[] args)
          int value = 10000;
          Btree theTree = new Btree(value);
          for(int j=0; j<value; j++)
               float n = (int) (java.lang.Math.random() *99);
               theTree.insert(n);
               theTree.display();
}

Binary Tree Question

Ok, I am making a binary tree. I have a question for my insert method. Firstly, I can't find out why the root node is inserted more than once in the tree. Also, I am having trouble with connecting the nodes that I insert to the tree. I can attach modes to root just fine, but I can't find out how to attach nodes to the existing nodes that are already attached to the tree. When I insert a node that meets the criteria of an already existing node, it replaces the node instead of getting attached to it. The answer is probably trivial, but I can't find it. Here is my insert method.
public void insert(T obj) {
          int _result1;
          int _result2;
     //TODO: Implement Q1 here
          if(isEmpty() == true){
               _root = createNode(obj, null, null, null);
               System.out.println("The root inserted is " + _root.element());
               insert(obj);
          _node = createNode(obj, _root, null, null);
          _result1 = _node.element().compareTo(_root.element());
          if(_result1 < 0){
                    if(_node.isInternal() == true){
                         _current = (BTreeNode<T>) _node2.element();
                         _result2 = _current.element().compareTo(_current.getParent().element());
                         //System.out.println("The current element is " + _current.element());
                         if(_result2 < 0){
                              _node1.getLeft();
                              _current = _node1.getLeft();
                         else{
                              _node1.getRight();
                              _current = _node1.getRight();
                         if(_node1.hasLeft() == true){
                              _node1.getLeft();
                              _current = _node1.getLeft();
                         else{
                              _current.setLeft(_node1);
                              _current = _node2;
                    else{
                         _node1 = createNode(obj, _node, null, null);
                         _node.setLeft(_node1);
                         _node1.setParent(_node);
                         System.out.println("The parent of the left node " + _node.element() + " is " + _node.getParent().element());
                         _root.setLeft(_node1);
                         _current = _node1;
                         //System.out.println("The current element is " + _node.element());
          else{
                    if(_node.isInternal() == true){
                         _current = (BTreeNode<T>) _current.element();
                         _result2 = _current.element().compareTo(_current.getParent().element());
                         //System.out.println("The current element is " +_current.element());
                         if(_result2 < 0){
                              _node1.getLeft();
                              _current = _node1.getLeft();
                         else{
                              _node1.getRight();
                              _current = _node1.getRight();
                         if(_node1.hasRight() == true){
                              _node1.getRight();
                              _current = _node1.getRight();
                         else{
                              _current.setLeft(_node1);
                              _current = _node1;
               else{
                    _node1 = createNode(obj, _node, null, null);
                    _node.setRight(_node1);
                    _node1.setParent(_node);
                    //_current = _node1;
                    System.out.println("The parent of the right node " + _node.element() + " is " + _node.getParent().element());
                    _root.setRight(_node1);
                    _current = _node1;
                    //System.out.println("The current element is " + _current.element());
          }The output I get is:
The root inserted is 6
The parent of the right node 6 is 6
The parent of the right node 6 is 6 ** I can't figure out why the root is inserted two extra times**
The parent of the left node 3 is 6
The parent of the right node 11 is 6
The parent of the right node 12 is 6 ** this node should be attaching to 11 instead of replacing it **
The parent of the right node 8 is 6
The parent of the right node 9 is 6
The parent of the right node 10 is 6
The parent of the right node 7 is 6
preorder :(6 (3 )(7 ))
postorder:(( 3) ( 7) 6)

IMO, your insert method is way too complicated.
Have a look at this pseudo code: that's all it takes to insert nodes in a BT:
class Tree<T extends Comparable<T>> {
    private Node root;
    public void insert(T obj) {
        'newNode' <- a new Node('obj') instance
        IF 'root' equals null
            let 'root' be the 'newNode'
        ELSE
            insert('root', 'newNode')
        END IF
    public void insert(Node parent, Node newNode) {
        IF 'parent' is less than 'newNode'
            IF the left child of 'parent' is null
                let 'newNode' be the left child of 'parent'
            ELSE
                make a recursive call here: insert('???', 'newNode')
            END IF
        ELSE
            IF the right child of 'parent' is null
                let 'newNode' be the right child of 'parent'
            ELSE
                make a recursive call here: insert('???', 'newNode')
            END IF
        END IF
}

Creating non binary trees in Java

Hi everybody actually i am used in creating binary trees in java,can u please tell me how do we create non binary -trees in java.

Hi,
Can u let us know the application area for Non binary tree.
S Rudra

Non binary trees in java

Hi guys i am used in creating binary trees ,tell me how do we create non binary trees in java.

public class Node {
private final Object payload;
private final Set<Node> children;
public Node(Object payload) {
this.payload = payload;
children = new HashSet<Node>();
// now add methods to add/remove children, a method to do something with the payload (say,
// a protected method that passes the payload to a method specified by some kind of interface),
// and methods to recurse over children.
}Actually rather than using Object probably should have generic-ized the class, but whatever.

Binary Tree in Java - ******URGENT********

HI,
i want to represent a binary tree in java. is there any way of doing that.
thanx
sraphson

HI,
i want to represent a binary tree in java. is there
e any way of doing that.
thanx
sraphsonFirst, what is a binary tree? Do you know how the binary tree looks like on puesdo code? How about a representation in terms of numbers? What is a tree? What is binary? The reason I ask is, what do you know about programming?
Asking to represent a binary tree in java seems like a question who doesn't know how it looks like in the first please. Believe me, I am one of them. I am not at this level yet. If you are taking a class that is teaching binary trees and you don't know how it looks like, go back to your notes.
Sounds harsh, but it is better to hear it from a person that doesn't know either then a boss that hired you because Computer Science was what you degree said. Yet, you don't know how to program?
Telling you will not help you learn. I can show you tutorials of trees would be start on where to learn.
HOW TO USE TREES (oops this is too simple, but it is a good example)
http://java.sun.com/docs/books/tutorial/uiswing/components/tree.html
CS312 Data Structures and Analysis of Algorithms
(Here is a course about trees. Search and learn)
http://www.calstatela.edu/faculty/jmiller6/cs312-winter2003/index.htm

Binary trees in Java

Hi there
Do you have any suggestions about how to implement trees or binary trees in Java?
As far as I know there are already some libraries for that, but I don't know how to use them. Was trying to get some implementation examples but I'm still very confused.
How can I use for example:
removeChild(Tree t)
addChild(Tree t)
isLeaf()
Thanks in advance

Lulu wrote:
Hi there
I have several questions about binary trees
Let's see, I use TreeMap to create them with the following code:
TreeMap treeMap = new TreeMap<Integer, String>();
treeMap.put("first", "Fruit");
treeMap.put("second","Orange");
treeMap.put("third", "Banana");
treeMap.put("fourth", "Apple");You've defined the map to hold integer keys and strings as values, yet you're trying to add string keys and string values to it: that won't work.
If this is a map how do I define if the data should go to the left or to the right of certain node?That is all done for you. In a TreeMap (using the no-args constructor), you can only store objects that are comparable to each other (so, they must implement the Comparable interface!). So the dirty work of deciding if the entry should be stored, or traversed, into the left or right subtree of a node, is all done behind the scenes.
Also note that TreeMap is not backed up by a binary tree, or a binary search tree, but by a red-black tree. A red-black tree is a self balancing binary tree structure.
Should I have dynamical keys so that they increase automatically when adding new data?
According to a webpage I should use Comparator(), is that to find the data in the tree and retrieve the key?
ThanksI am not sure what it is you want. I am under the impression that you have to write a binary tree (or a binary search tree) for a course for school/university. Is that correct? If so, then I don't think you're permitted to use a TreeMap, but you'll have to write your own classes.

Drawing a binary tree

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