Binary Tree Postorder Traversal

一、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]

.
二、思路

二叉树的后续遍历，非递归遍历稍微复杂一点，需要另外设置一个数据结构，改天补上，未完待续。

三、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:
void postorder(TreeNode* node,vector<int> &res){
if(!node){
return ;
}else{
postorder(node -> left,res);
postorder(node -> right,res);
res.push_back(node -> val);
}
}
vector<int> postorderTraversal(TreeNode* root) {
vector<int> vec;
postorder(root,vec);
return vec;
}
};