文章标题

1. 问题描述
   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].

2. 解决思路
   其实后序遍历可以转换成前序遍历：先遍历根节点，然后遍历右子树，然后遍历左子树，然后得到的结果反转一下就可以了。

3. 代码

/** * Definition for a binary tree node. * 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) {        if (!root)            return vector<int>();        stack<TreeNode*> st;        vector<int> res;        st.push(root);        TreeNode *pre;        while(!st.empty()) {            pre = st.top();            st.pop();            res.push_back(pre->val);            if (pre->left)                st.push(pre->left);            if (pre->right)                st.push(pre->right);        }        reverse(res.begin(),res.end());        return res;    }};