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。