Given a Binary Tree, print the nodes of a binary tree in in-order fashion.
In the in-order traversal for a given node 'n',
1. We first traverse left-subtree of 'n' by calling in_order_print(n.left)
2. Then we visit node 'n' itself.
3. And finally we traverse right-subtree of 'n' by calling in_order_print(n.right)
Please check out the code below(in_order_print()) with visualization for illustration.
package com.ideserve.nilesh.questions;
public class Tree {
private TreeNode root;
public TreeNode getRoot() {
return root;
}
class TreeNode {
private int data;
private TreeNode left;
private TreeNode right;
public TreeNode(int data, TreeNode left, TreeNode right) {
super();
this.data = data;
this.left = left;
this.right = right;
}
public int getData() {
return data;
}
public void setData(int data) {
this.data = data;
}
public TreeNode getLeft() {
return left;
}
public void setLeft(TreeNode left) {
this.left = left;
}
public TreeNode getRight() {
return right;
}
public void setRight(TreeNode right) {
this.right = right;
}
public TreeNode() {
super();
}
public TreeNode(int data) {
super();
this.data = data;
}
@Override
public String toString() {
return data+"";
}
}
public static void main(String[] args) {
Tree tree = new Tree();
tree.createSampleTree();
tree.printInorder();
}
/*
* Create a sample tree
* 1
* 2 3
* 4 5 6 7
*
*/
public void createSampleTree() {
root = new TreeNode(1, new TreeNode(2, new TreeNode(4), new TreeNode(5)), new TreeNode(3, new TreeNode(6), new TreeNode(7)));
}
/*
* Print inorder traversal
*/
public void printInorder() {
printInorder(root);
}
private void printInorder(TreeNode root) {
if(root == null) {
return;
}
printInorder(root.getLeft());
System.out.print(root.getData() + " ");
printInorder(root.getRight());
}
}
Time Complexity is O((n))
Space Complexity is O((1))