leetcode 145. 二叉树的后序遍历

给定一个二叉树，返回它的 后序 遍历。

示例:

输入: [1,null,2,3]
   1
         2
    /
   3 

输出: [3,2,1]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     void dfs(TreeNode* root, vector<int>& ans){
13         if(root == NULL) return;
14         if(root->left) dfs(root->left, ans);
15         if(root->right) dfs(root->right, ans);
16         ans.push_back(root->val);
17     }
18     vector<int> postorderTraversal(TreeNode* root) {
19         vector<int> ans;
20         dfs(root, ans);
21         return ans;
22     }
23 };

迭代的实现：用栈的方式来实现

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 #include<stack>
11 class Solution {
12 public:
13     vector<int> postorderTraversal(TreeNode* root) {
14         stack<TreeNode*> s;
15         s.push(root);
16         vector<int> ans;
17         if(root == NULL) return ans;
18         while(!s.empty()){
19             TreeNode* temp = s.top();
20             if(temp->left){
21                 s.push(temp->left);
22                 temp->left = NULL;
23             } else if(temp->right){
24                 s.push(temp->right);
25                 temp->right = NULL;
26             } else{
27                 ans.push_back(temp->val);
28                 s.pop();
29             }
30         }
31         return ans;
32     }
33 };

原文地址：https://www.cnblogs.com/mr-stn/p/8978055.html