How to Pretty Print a Binary Tree?

I'm trying to display a Binary Tree in such a way:
________26
___13_________2
1_______4 3_________1
(without the underscores)
however I cannot figure out the display method.
class BinaryNode
          //Constructors
          BinaryNode leftChild, rightChild;
Object data;
     BinaryNode()
leftChild = null;
data = null;
rightChild = null;
          BinaryNode( Object d, BinaryNode left, BinaryNode right)
leftChild = left;
data = d;
rightChild = right;
          //Height
          public static int Height(BinaryNode root)
if (root == null)
return 0;
if ((Height(root.leftChild)) > Height(root.rightChild))
return 1 + Height(root.leftChild);
return 1 + Height(root.rightChild);
          //Count
public static int Count(BinaryNode root)
if(root==null)
return 0;
return 1 + Count(root.leftChild) + Count(root.rightChild);
          //Display
          public static void Display(BinaryNode root)
          int level = 2^(Level(root)-1)
          for (int i = 1; i<Height(root)+1; i++)
          System.out.printf("%-4s%
          Display(root, i);
          System.out.println();
          public static void Display(BinaryNode root, int level)
          if (root!=null)
          if(level==1)
          System.out.print(root.data + " ");
          else
          Display(root.leftChild, level-1);
          Display(root.rightChild, level-1);
          //Level
          public static int Level(BinaryNode root)
          if(root==null)
          return 0;
          if(root.leftChild == null && root.rightChild == null)
          return 1;
          return Level(root.leftChild) + Level(root.rightChild);
Edited by: 815035 on Nov 23, 2010 12:27 PM

The example of what the OP wants it to look like I thought was quite plain. Its right at the top of the post.
Unfortunately it is also quite difficult to accomplish using System.out.print statements.
You have to print out the root of the tree first (its at the top)
However you don't know how far along to the right you need to print it without traversing the child nodes already (you need to know how deep the tree is, and how far to the left the tree extends from the root)
So you will need to traverse the tree at least twice.
Once to work out the offsets, and again to print out the values.
The working out of offsets would have to be a depth search traversal I think
The printing of the values in this fashion would be a breadth first traversal.
I remember (ages ago) doing a similar assignment, except we printed the tree sideways.
ie the root was on the left, the leaves of the tree on the right of the screen.
That meant you could do an inorder depth traversal of the tree to just print it once.
hope this helps,
evnafets

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The example of what the OP wants it to look like I thought was quite plain. Its right at the top of the post.
Unfortunately it is also quite difficult to accomplish using System.out.print statements.
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I remember (ages ago) doing a similar assignment, except we printed the tree sideways.
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hope this helps,
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public class Node {
private final Object payload;
private final Set<Node> children;
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// now add methods to add/remove children, a method to do something with the payload (say,
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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?!?
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Code:
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abstract double evaluate(VariableInput v);
abstract String print();
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Code:
public class UnaryNode extends Node {
private UnaryFunction operator;
private Node left;
public UnaryNode(UnaryFunction op, Node terminal) {
operator = op;
this.left = terminal;
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void setLeft(Node left) {
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Node getLeft() {
return left;
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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.
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Just having a quick look over it, and it looks like you are trying to add an extra argument in the TreeNode() method (unless there is a bit of overloading and there is a second treenode method in there) As it is TreeNode only accepts two argumets you have 3
As for printing the tree you would need to flatten it, that is an in order traversal of the tree.
FWIW
I just finished a project at uni that involved at frist writing a BST and then an AVL tree. the full point of these things seems to be to keep students awake at night*
*Before anyone flames, it's a joke
G

How to remember the path while traverse a binary tree?

Hi, again, I have difficulty handling tree search problems.
The quesion is How to search for a node from a binary tree A, return true if found meanwhile generate the path which can be used to locate the node.
I think the signature should be:
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public staic boolean search(Node rootOfA, Node b, ArrayList<Integer> path){
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Node{
Node left;
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I know if there is another field in the Node class (say, a flag), the quesion would be much easier, but the space is really critical.
I tried to use recursion but havn't got a correct solution. I am thinking of usinga non-recursion algo.
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Hi, JosAh,
That's mind provoking. However, I do think it works. It does not pop some stuff it should pop. I tested it over a very simple tree and it failed. Did you test it? I might be wrong...
The tree I am working on does not have null pointers, the condition to test if a node is a leaf is if(node.right == right). Namly, all the right pointer of leaves points to itself.
So I changed your code to be:
Stack search(Node root, Node node, Stack<Integer> path) {
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if (search(root.left, node, path) != null) return path;
path.pop();
path.push(1);
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}I simply tested it with
Stack<Integer> path = new Stack<Integer>();
System.out.println( root, root.right.right, path);
root is the root of a complete binary tree with 7 nodes(4 leaves).
Apparenly, if the path is built correctly search(root, root.right.right, path) would return [1,1] whereas this seach returns [ 0 , 1, 1].
Considerring , the right branch never pops, I changed it into
Then I changed it to :
Stack search(Node root, Node node, Stack<Integer> path) {
     if (root == null || root.right ==right ) return null;
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     path.push(0);
if (search(root.left, node, path) != null) return path;
path.pop();
path.push(1);
if (search(root.right, node, path) != null) return path;
path.pop();
return path;
With the same test case, it returns [].
I will keep working on it.
Cheers,
Message was edited by:
since81
Message was edited by:
since81

How to make a binary tree

i have this assignment for school. The first thing it says is: Write a binary tree class named my_Tree as the basic data structure. What does this mean? Can anyone give me an example of the java code that this should be written in?
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i dont really understand that? Im really slow at
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// copywrong filestream. no rights reserved
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How to serialize a binary tree object

HI Friends,
This is the problem.............................
I got a binary Tree Program with root (object ) containing the elements in the tree.
I got a server which performs insertions and deletions in the root.
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Have you looked at the TreeSet code.
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How do we know if it's a leaf in a binary tree

how do we know if a node is a leaf in a binary tree?

Both the left and right nodes should be null.

How to extend breadth first Search for Binary Tree to any kind of Tree??

Dear Friends,
I am thinking a problem, How to extend breadth first Search for Binary Tree to any kind of Tree?? ie each node has more than 2 leaves such as 1, 2,3,4 or any,
I have following code to successfully apply for breadth first Search in Binary Tree as follows,
package a.border;
import java.util.ArrayList;
import java.util.LinkedList;
public class Tree
    int root;
    Tree left;
    Tree right;
    static ArrayList<Integer> list = new ArrayList<Integer>();
    static ArrayList<Tree> treeList = new ArrayList<Tree>();
    private static LinkedList<Tree> queue = new LinkedList<Tree>();
     * @param root root value
     * @param left left node
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        this.right = right;
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     * You really should know what this does...
     * @param root
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        this.root = root;
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        if (left != null)
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here this code is for the tree like:
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         *          t =           2       6
         *                      1   3    5   7
         */But i hope the new code can handle tree like:
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         *                           /   | \
         *                          /     |   \
         *          t =            2     8   6
         *                        / | \    |    /| \
         *                      1 11 3 9   5 10 7
         */Thanks

sunnymanman wrote:
Dear Friends,
I am thinking a problem, How to extend breadth first Search for Binary Tree to any kind of Tree?? ...The answer is interfaces.
What do all trees have in common? And what do all nodes in trees have in common?
At least these things:
interface Tree<T> {
    Node<T> getRoot();
interface Node<T> {
    T getData();
    List<Node<T>> getChildren();
}Now write concrete classes implementing these interfaces. Let's start with a binary tree (nodes should have comparable items) and an n-tree:
class BinaryTree<T extends Comparable<T>> implements Tree<T> {
    protected BTNode<T> root;
    public Node<T> getRoot() {
        return root;
class BTNode<T> implements Node<T> {
    private T data;
    private Node<T> left, right;
    public List<Node<T>> getChildren() {
        List<Node<T>> children = new ArrayList<Node<T>>();
        children.add(left);
        children.add(right);
        return children;
    public T getData() {
        return data;
class NTree<T> implements Tree<T> {
    private NTNode<T> root;
    public Node<T> getRoot() {
        return root;
class NTNode<T> implements Node<T> {
    private T data;
    private List<Node<T>> children;
    public List<Node<T>> getChildren() {
        return children;
    public T getData() {
        return data;
}Now with these classes, you can wite a more generic traversal class. Of course, every traversal class (breath first, depth first) will also have something in common: they return a "path" of nodes (if the 'goal' node/data is found). So, you can write an interface like this:
interface Traverser<T> {
    List<Node<T>> traverse(T goal, Tree<T> tree);
}And finally write an implementation for it:
class BreathFirst<T> implements Traverser<T> {
    public List<Node<T>> traverse(T goal, Tree<T> tree) {
        Node<T> start = tree.getRoot();
        List<Node<T>> children = start.getChildren();
        // your algorithm here
        return null; // return your traversal
}... which can be used to traverse any tree! Here's a small demo of how to use it:
public class Test {
    public static void main(String[] args) {
        Tree<Integer> binTree = new BinaryTree<Integer>();
        // populate your binTree
        Tree<Integer> nTree = new NTree<Integer>();
        // populate your nTree
        Traverser<Integer> bfTraverser = new BreathFirst<Integer>();
        // Look for integer 6 in binTree
        System.out.println("bTree bfTraversal -> "+bfTraverser.traverse(6, binTree));
        // Look for integer 6 in nTree
        System.out.println("bTree bfTraversal -> "+bfTraverser.traverse(6, nTree));
}Good luck!

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.
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For comparison purpose, the same random numbers are stored and sorted in an internal table and printed.
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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.
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exporting
rnd_min = 0
rnd_max = 100
importing
rnd_value = int.
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rnd-num = int.
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perform add_value using tree int.
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skip.
loop at rnd.
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endloop.
skip.
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skip.
perform print_value using tree.
skip.
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-->VAL text
form add_value using tree type stree val type i.
field-symbols: <ltree> type any.
data: work type stree.
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tree-value = val. " assign value
clear: tree-left, tree-right.
create data tree-left type stree. "Create an empty node for left
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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
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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
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endif.
endform. "print_value
pls reward if helps,
regards.

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