Binary Tree InOrder traversal in java

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This is 4th part of java binary tree tutorial.

Table of Contents

InOrder traversal:
- Recursive solution
- Iterative solution
Java Binary tree tutorial
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In this post, we will see about InOrder binary tree traversal in java.

InOrder traversal:

In InOrder traversal,each node is processed between subtrees.In simpler words,Visit left subtree, node and then right subtree.

Steps for InOrder traversal are:

Traverse the left subtree in InOrder.
Visit the node.
Traverse the right subtree in InOrder.

There can be two ways of implementing it

Recursive
Iterative

Recursive solution

Recursive solution is very straight forward.Below diagram will make you understand recursion better.

Code for recursion will be:

// Recursive Solution

public void inOrder(TreeNode root) {

if(root != null) {

inOrder(root.left);

//Visit the node by Printing the node data

System.out.printf("%d ",root.data);

inOrder(root.right);

}

Iterative solution

For recursion, we use implicit stack. So here to convert recursive solution to iterative, we will use explicit stack.
Steps for iterative solution:

Create an empty stack s and Initialize currentNode as root
Push the currentNode to s and set currentNode = currentNode.left until currentNode is NULL
If currentNode is NULL and s is not empty then
- Pop the top node from stack s and print it
- set currentNode = currentNode.right
- go to step 2
If stack is empty and currentNode is also null then we are done with it

// Iterative solution

public void inOrderIter(TreeNode root) {

if(root == null)

return;

Stack<TreeNode> s = new Stack<TreeNode>();

TreeNode currentNode=root;

while(!s.empty() || currentNode!=null){

if(currentNode!=null)

{

s.push(currentNode);

currentNode=currentNode.left;

}

else

{

TreeNode n=s.pop();

System.out.printf("%d ",n.data);

currentNode=n.right;

}

Lets create java program for InOrder traversal:

package org.arpit.java2blog.binarytree;

import java.util.Stack;

public class BinaryTreeInOrder {

public static class TreeNode

{

int data;

TreeNode left;

TreeNode right;

TreeNode(int data)

{

this.data=data;

}

// Recursive Solution

public void inOrder(TreeNode root) {

if(root != null) {

inOrder(root.left);

//Visit the node by Printing the node data

System.out.printf("%d ",root.data);

inOrder(root.right);

}

// Iterative solution

public void inOrderIter(TreeNode root) {

if(root == null)

return;

Stack<TreeNode> s = new Stack<TreeNode>();

TreeNode currentNode=root;

while(!s.empty() || currentNode!=null){

if(currentNode!=null)

{

s.push(currentNode);

currentNode=currentNode.left;

}

else

{

TreeNode n=s.pop();

System.out.printf("%d ",n.data);

currentNode=n.right;

}

public static void main(String[] args)

{

BinaryTreeInOrder bi=new BinaryTreeInOrder();

// Creating a binary tree

TreeNode rootNode=createBinaryTree();

System.out.println("Using Recursive solution:");

bi.inOrder(rootNode);

System.out.println();

System.out.println("-------------------------");

System.out.println("Using Iterative solution:");

bi.inOrderIter(rootNode);

}

public static TreeNode createBinaryTree()

{

TreeNode rootNode =new TreeNode(40);

TreeNode node20=new TreeNode(20);

TreeNode node10=new TreeNode(10);

TreeNode node30=new TreeNode(30);

TreeNode node60=new TreeNode(60);

TreeNode node50=new TreeNode(50);

TreeNode node70=new TreeNode(70);

rootNode.left=node20;

rootNode.right=node60;

node20.left=node10;

node20.right=node30;

node60.left=node50;

node60.right=node70;

return rootNode;

}

Run above program and you will get following output:

Using Recursive solution:
10 20 30 40 50 60 70
————————-
Using Iterative solution:
10 20 30 40 50 60 70

Java Binary tree tutorial

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Binary Tree InOrder traversal in java