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

Example:

Input: [1,null,2,3]   1    \     2    /   3Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?

难度：medium

题目：给定二叉树，返回其中序遍历结点。（不要使用递归）

思路：栈

Runtime: 1 ms, faster than 55.50% of Java online submissions for Binary Tree Inorder Traversal.

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public List
    
      inorderTraversal(TreeNode root) {        Stack
     
       stack = new Stack<>();        List
      
        result = new ArrayList<>();        while (!stack.isEmpty() || root != null) {            if (root != null) {                stack.push(root);                root = root.left;            } else {                TreeNode node = stack.pop();                result.add(node.val);                root = node.right;            }        }                return result;    }}