leetcode Binary Tree Postorder

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

题意：后序不让用递归遍历二叉树

思路：先root进站，取栈顶

1，一直找到最左的子节点，边找边进站，成为新的栈顶

2.判断栈顶的右子节点有没有被访问

3.如果没有，取右子节点，重复1的工作

4，如果有，按照后序遍历，既然右边都被访问了，那么肯定左边在右边之前被访问了，那么就可以访问本身节点了，然后向上走一级，把刚才访问的节点当成上一级的右子节点

代码：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<int> postorderTraversal(TreeNode *root) {        stack<TreeNode*> stack1;        vector <int> result;        bool visited;        TreeNode *p = root,*q;        if(!root) return result;        do{            while(p!=NULL){                stack1.push(p);                p = p->left;            }            visited = true;            q = NULL;            while(!stack1.empty() && visited){                p = stack1.top();                if(p ->right ==q){                    result.push_back(p->val);                    stack1.pop();                    q = p;                }                else{                    p = p->right;                    visited =false;                }            }        }while(!stack1.empty());        return result;    }};