94. Binary Tree Inorder Traversal

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

For example:
Given binary tree [1,null,2,3],

   1    \     2    /   3

return [1,3,2].

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

中序遍历，参照模板：JAVA 二叉树遍历。代码如下：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> inorderTraversal(TreeNode root) {        List<Integer> res = new ArrayList<Integer>();        Stack<TreeNode> stack = new Stack<TreeNode>();          TreeNode node = root;        while (node != null || stack.size() > 0) {            while (node != null) {                stack.push(node);                node = node.left;            }            if (stack.size() > 0) {                node = stack.pop();                res.add(node.val);                node = node.right;            }        }        return res;    }}