Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes’ values.

Example :
Given binary tree

   1
    \
     2
    /
   3

return [1,3,2].

Using recursion is not allowed.

Solution:

Time: O(n)
Space: O(n)

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) {
 *      val = x;
 *      left=null;
 *      right=null;
 *     }
 * }
 */
public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode A) {
        ArrayList<Integer> result = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode curr = A;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                stack.push(curr);
                curr = curr.left;
            }
            if (!stack.isEmpty()) {
                curr = stack.pop();
                result.add(curr.val);
                curr = curr.right;
            }
        }
        return result;
    }
}