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package tree;

import java.util.ArrayList;

import java.util.List;

import java.util.Stack;

/**

* Created by gouthamvidyapradhan on 06/08/2017. Given a binary tree, return the inorder traversal

* of its nodes' values.

*

* <p>For example: Given binary tree [1,null,2,3], 1 \ 2 / 3 return [1,3,2].

*

* <p>Note: Recursive solution is trivial, could you do it iteratively?

*/

public class BinaryTreeInorderTraversal {

public static class TreeNode {

int val;

TreeNode left;

TreeNode right;

TreeNode(int x) {

val = x;

}

}

public static void main(String[] args) throws Exception {

TreeNode root = new TreeNode(3);

root.left = new TreeNode(4);

root.left.left = new TreeNode(5);

root.left.right = new TreeNode(6);

root.left.left.left = new TreeNode(9);

root.left.left.right = new TreeNode(10);

root.right = new TreeNode(2);

root.right.left = new TreeNode(7);

root.right.right = new TreeNode(8);

List<Integer> result = new BinaryTreeInorderTraversal().inorderTraversal(root);

System.out.println(result);

}

public List<Integer> inorderTraversal(TreeNode root) {

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

TreeNode curr = root;

List<Integer> result = new ArrayList<>();

while (curr != null || !stack.isEmpty()) {

while (curr != null) {

stack.push(curr);

curr = curr.left;

}

curr = stack.pop();

result.add(curr.val);

curr = curr.right;

}

return result;

}

}