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
Similar Messages
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Hi guys ;
I have a BST class .One of its methods is preorder travesal.
this method traverse theBinary tree in preorder and prints nodes data to a file.
Now what i want to do is to be able to construct the binary tree by reading data from this file : keeping in mind the data nodes were saved to this file in preorder.
Is this a possible thing to do ?
How can I do this ?
Is there a better way to save a binary tree to a file and the other way : read from file and construct the tree again?
Many thanks for helping !I'd like to say thanks too.. it really helped me out!
I used a few minutes (or so ;) to figure out the functions for saving and rebuilding the tree, so I thought I'd post it here, if anyone could use it. Note: this ugly pseudocode might look a little like VB.net-code, which happened completely by accident..
Private Function SearchForLeaves(ByVal StartNode As Node, ByVal BitString As String) As String
'Recursive function. If a leaf is found on left or right side of the branch,
'then print the binary pattern used to get there, else call the function again on the next branch.
Dim SavedTree As String
With StartNode
If .LeftChild.Type = NodeType.Leaf Then
Print .leftChild.Value & " " & Bitstring & "0"
SavedTree += "!" & .LeftChild.Value
Else
SearchForLeaves(.LeftChild, BitString & "0")
End If
If .RightChild.Type = NodeType.Leaf Then
Print .leftChild.Value & " " & Bitstring & "1"
SavedTree += "!" & .RightChild.Value
Else
SearchForLeaves(.RightChild, BitString & "1")
End If
SavedTree += "?"
End With
Return SavedTree
End Function
Private Function RebuildTree(ByVal savedTree As String) As Node
Dim n As Node
While Len(savedTree) > 0
If savedTree.StartsWith("!") Then
CropString(savedTree)
n = New Node(NodeType.Leaf)
n.Value = CropString(savedTree)
Stack.Push(n)
End If
If savedTree.StartsWith("?") Then
CropString(savedTree)
n = New Node(NodeType.Branch)
n.RightChild = Stack.Pop
n.LeftChild = Stack.Pop
n.LeftChild.Parent = n
n.RightChild.Parent = n
Stack.Push(n)
End If
End While
Return Stack.Pop
End Function
Private Function CropString(byRef cropMe as String) As String
Dim L As String
L = cropMe.Substring(0, 1)
cropMe = cropMe.Substring(1)
Return L
End FunctionCropString just crops off and returns the first character of the string.. I'm going to rewrite this whole thing, but right now I'm just pleased I figured out how to make it work :)
Cheers! -
Building a binary tree from a string
I'm having trouble building a binary tree with shape described in a string.
00 - means both left and right subtrees are empty;
01 - means that the left is empty, and the right is not;
10 - right is empty, left is not
11 - neither are empty
public BinaryTree(String s, Iterator it) {
This constructor is supposed to build the binary tree of the specified shape and fills it with values. THe values come from Iterator it, and are placed into the tree in preorder fashion.
I have to complete this constructor but I don't really know what to do. I was thinking I could use indexOf somehow. Any help would be greatly appreciated.
Thanks,
MikeI'd build it like this (this is from the top of my head, so no typo-free-warranties etc.) -- public class Tree {
private Object data;
private Tree left= null;
private Tree right= null;
private Tree(StringReader sr, Iterator di) {
char l= (char)sr.read(); // left and right subtree indicators
char r= (char)sr.read();
data= di.next(); // set data for this node
if (l == '1') left= new Tree(sr, di); // build left subtree
if (r == '1') right= new Tree(sr, di); // build right subtree
public Tree(String s, Iterator di) {
this(new StringReader(s), di);
} Note that the private constructor (the one that does all the work) doesn't handle incorrect strings
at all, i.e. it'll crash horribly when the string passes contains, say an odd number of characters
is is simply passing incorrect construction information. Also note that the Iterator must be able
to iterate over enough elements to be set.
kind regards,
Jos -
So I have an assignment to write a program that deciefers basic morse code using a binary tree. A file was provided with the information for the binary tree and I am using the text books author's version of BinaryTree class with a few methods of my own thrown in.
For starters .. why would you ever use a tree for this .. just make a look up table. But my real problem is i keep getting null pointers at 38, 24, 55 and I have tried just about everything I know to do at this point. My mind is exhausted and I think I need a fresh pair of eyes to tell me what I have done wrong.
I am sorry for such sloppy code.. I have been changing and rearranging too much.
import java.io.*;
import java.util.*;
public class MorseCode extends BinaryTree{
static BufferedReader in;
static BinaryTree<String> bT = new BinaryTree<String>();
/* public static void loadTree() throws FileNotFoundException, IOException{
in = new BufferedReader(new FileReader("MorseCode.txt"));
bT = readBinaryTree(in);
public static String decode(Character c) throws IOException{
if(c.equals("null")){
return "";
}else if(c.equals(" ")){
return " ";
}else{
return (":" + find(c));
public static String find(Character c) throws IOException, FileNotFoundException{
in = new BufferedReader(new FileReader("MorseCode.txt"));
bT = readBinaryTree(in);
Queue<BinaryTree> data = new LinkedList<BinaryTree>();
BinaryTree<String> tempTree = bT;
String temp = null;
Character tempChar = null;
data.offer(tempTree);
while(!data.isEmpty()){
tempTree = data.poll();
temp = tempTree.getData();
tempChar = temp.charAt(0);
if(tempChar.equals(c)){
break;
data.offer(tempTree.getRightSubtree());
data.offer(tempTree.getLeftSubtree());
return temp.substring(2);
public static void main(String[] args) throws FileNotFoundException, IOException{
Scanner scan = new Scanner(new FileReader("encode.in.txt"));
String s = "";
String temp = "";
while(scan.hasNextLine()){
temp = scan.nextLine();
for(int i = 0; i < temp.length(); i++){
s = s + decode(temp.charAt(i));
System.out.println(s);
/** Class for a binary tree that stores type E objects.
* @author Koffman and Wolfgang
class BinaryTree <E> implements Serializable
//===================================================
/** Class to encapsulate a tree node. */
protected static class Node <E> implements Serializable
// Data Fields
/** The information stored in this node. */
protected E data;
/** Reference to the left child. */
protected Node <E> left;
/** Reference to the right child. */
protected Node <E> right;
// Constructors
/** Construct a node with given data and no children.
@param data The data to store in this node
public Node(E data) {
this.data = data;
left = null;
right = null;
// Methods
/** Return a string representation of the node.
@return A string representation of the data fields
public String toString() {
return data.toString();
}//end inner class Node
//===================================================
// Data Field
/** The root of the binary tree */
protected Node <E> root;
public BinaryTree()
root = null;
protected BinaryTree(Node <E> root)
this.root = root;
/** Constructs a new binary tree with data in its root,leftTree
as its left subtree and rightTree as its right subtree.
public BinaryTree(E data, BinaryTree <E> leftTree, BinaryTree <E> rightTree)
root = new Node <E> (data);
if (leftTree != null) {
root.left = leftTree.root;
else {
root.left = null;
if (rightTree != null) {
root.right = rightTree.root;
else {
root.right = null;
/** Return the left subtree.
@return The left subtree or null if either the root or
the left subtree is null
public BinaryTree <E> getLeftSubtree() {
if (root != null && root.left != null) {
return new BinaryTree <E> (root.left);
else {
return null;
/** Return the right sub-tree
@return the right sub-tree or
null if either the root or the
right subtree is null.
public BinaryTree<E> getRightSubtree() {
if (root != null && root.right != null) {
return new BinaryTree<E>(root.right);
} else {
return null;
public String getData(){
if(root.data == null){
return "null";
}else{
return (String) root.data;
/** Determine whether this tree is a leaf.
@return true if the root has no children
public boolean isLeaf() {
return (root.left == null && root.right == null);
public String toString() {
StringBuilder sb = new StringBuilder();
preOrderTraverse(root, 1, sb);
return sb.toString();
/** Perform a preorder traversal.
@param node The local root
@param depth The depth
@param sb The string buffer to save the output
private void preOrderTraverse(Node <E> node, int depth,
StringBuilder sb) {
for (int i = 1; i < depth; i++) {
sb.append(" ");
if (node == null) {
sb.append("null\n");
else {
sb.append(node.toString());
sb.append("\n");
preOrderTraverse(node.left, depth + 1, sb);
preOrderTraverse(node.right, depth + 1, sb);
/** Method to read a binary tree.
pre: The input consists of a preorder traversal
of the binary tree. The line "null" indicates a null tree.
@param bR The input file
@return The binary tree
@throws IOException If there is an input error
public static BinaryTree<String> readBinaryTree(BufferedReader bR) throws IOException
// Read a line and trim leading and trailing spaces.
String data = bR.readLine().trim();
if (data.equals("null")) {
return null;
else {
BinaryTree < String > leftTree = readBinaryTree(bR);
BinaryTree < String > rightTree = readBinaryTree(bR);
return new BinaryTree < String > (data, leftTree, rightTree);
}//readBinaryTree()
/*Method to determine the height of a binary tree
pre: The line "null" indicates a null tree.
@param T The binary tree
@return The height as integer
public static int height(BinaryTree T){
if(T == null){
return 0;
}else{
return 1 +(int) (Math.max(height(T.getRightSubtree()), height(T.getLeftSubtree())));
public static String preOrderTraversal(BinaryTree<String> T){
String s = T.toString();
String s2 = "";
String temp = "";
Scanner scan = new Scanner(s);
while(scan.hasNextLine()){
temp = scan.nextLine().trim();
if(temp.equals("null")){
s2 = s2;
}else{
s2 = s2 + " " + temp;
return s2;
}//class BinaryTreeAs well as warnerja's point, you say you keep getting these errors. Sometimes it's helpful when illustrating a problem to replace the file based input with input that comes from a given String. That way we all see the same behaviour under the same circumstances.
public static void main(String[] args) throws FileNotFoundException, IOException{
//Scanner scan = new Scanner(new FileReader("encode.in.txt"));
Scanner scan = new Scanner("whatever");The following isn't the cause of an NPE, but it might be allowing one to "slip through" (ie you think you've dealt with the null case when you haven't):
public static String decode(Character c) throws IOException{
if(c.equals("null")){
c is a Character so it will never be the case that it is equal to the string n-u-l-l.
Perhaps the behaviour of each of these methods needs to be (documented and) tested. -
hello
I want to do a combat using a binary tree, the thing works as follows
for example given this tree
1
2 3
4 5 6 7
8 9 10 11I have figthers at node 4,5,8,9,10,11 all other nodes that are not initial are distances (1,2,3,6,7), then when to branches join there's a combat. I would like to have an idea on how to move elements throught this tree. I know many algorithms to run across the tree, but can't think on how to make them fight, I don't know how to relate it.
Any ideas are apreciated
thxs in advance :)I don't know if this will help you but let's say the top room is 1. To construct the room-number of the sub-room to the left you do 2*1 = 2, and to the right 2*1 + 1 = 3. If you are at any room say N you can construct the left sub-room number like N*2, and the right like N*2+1. This is basically just another way to represent the nodes of a binary tree, as numbers.
If you construct the room numbers like above, say you have two fighters in room 2 and 3. To get the common super-room you divide the numbers with 2, like 2/2=1 and 3/2 = 1. So the super-room of 2 and 3 is the same, namely 1. Divide any room-number with 2 and you get the super-room.
When the game start each fighter is given a room-number. If you divide all room-numbers with two this corresponds to each fighter going to a super-room. If two fighters end up with the same room-number, well then they're in the same room and must fight. -
Help needed in constructing a tree
Help needed in constructing a tree. I was wondering if some one can suggest me how to add messages in the second column for both the parent node and child elements.
I was able to create a tree succefully, but want to add some description in the second column for the first column elements, for both parent and child elements.
Please suggest me how to add the arrays to second column for parent and child nodes.
Solved!
Go to Solution.
Attachments:
Tree_fix.vi 15 KBThe Child Text parameter is the one you are searching for. It accepts a 1D string array for the following columns.
hope this helps,
Norbert
CEO: What exactly is stopping us from doing this?
Expert: Geometry
Marketing Manager: Just ignore it. -
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 TreeAppString 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 -
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:
since81I'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.
:) -
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();
} -
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
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