LeetCode-Binary Tree Postorder Traversal

题目：https://oj.leetcode.com/problems/binary-tree-postorder-traversal/

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

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [3,2,1].

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

源码：Java版本
算法分析：双栈。时间复杂度O(n)，空间复杂度O(n)

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> postorderTraversal(TreeNode root) {        List<Integer> result=new ArrayList<Integer>();          Stack<TreeNode> stack=new Stack<TreeNode>();         Stack<Integer> cntStack=new Stack<Integer>();        TreeNode node=root;          int label;        do {            while(node!=null) {                stack.push(node);                cntStack.push(0);                node=node.left;            }            if(!stack.isEmpty()) {                node=stack.pop();                label=cntStack.pop();                if(label==0) {                    cntStack.push(1);                    stack.push(node);                    node=node.right;                }else {                    result.add(node.val);                    node=null;  //import                }            }                    }while(!stack.isEmpty());                return result;    }}

代码解释：cntStack用来标志顶点是第一次还是第二次出栈。第一次出栈需要访问其由节点，第二次出栈直接访问它。