# Description

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].

The original problem is here.

The original code is here.

# My Solution

I solve this problem in C++, as below:

``````/*
*Binary Tree Postorder Traversal
*Author: shuaijiang
*Email: zhaoshuaijiang8@gmail.com
*/
#include<iostream>
#include<stack>
#include<vector>
#include<stdlib.h>
using namespace std;

/**
* 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) {
vector<int> vec;
stack<TreeNode*> myStack;
if(root==NULL)
return vec;

myStack.push(root);
while(!myStack.empty()){
TreeNode *node = myStack.top();
myStack.pop();
TreeNode *left = node->left;
TreeNode *right = node->right;
if(left == NULL && right == NULL){
vec.push_back(node->val);
}
else{
node->left = NULL;
node->right = NULL;
myStack.push(node);
}
if(right!=NULL)
myStack.push(right);
if(left!=NULL)
myStack.push(left);
}
return vec;
}
};
``````

# Note

To solve the problem, a stack is used. The postorder traversal of the binary tree is left, right and root. So that the order in the stack is root, right and left.