【树】145. 二叉树的后序遍历

题目：

解法:

方法一：递归实现

 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 class Solution {
11 public:
12     vector<int> postorderTraversal(TreeNode* root) 
13     {
14         vector<int> ret;
15         if (NULL == root)
16         {
17             return ret;
18         }
19 
20         postorderTraversalHelper(root, ret);
21 
22         return ret;
23     }
24 
25     void postorderTraversalHelper(TreeNode *root, vector<int> &ret)
26     {
27         if (NULL != root)
28         {
29             postorderTraversalHelper(root->left, ret);
30             postorderTraversalHelper(root->right, ret);
31             ret.push_back(root->val);
32         }
33     }
34 };

方法二：迭代实现

 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 class Solution {
11 public:
12     vector<int> postorderTraversal(TreeNode* root) 
13     {
14         vector<int> ret;
15         if (NULL == root)
16         {
17             return ret;
18         }
19 
20         stack<TreeNode *> st;
21         st.push(root);
22 
23         while (!st.empty())
24         {
25             TreeNode *tmp = st.top();
26             st.pop();
27             ret.insert(ret.begin(), tmp->val);
28 
29             if (tmp->left)
30             {
31                 st.push(tmp->left);
32             }
33 
34             if (tmp->right)
35             {
36                 st.push(tmp->right);
37             }
38         }
39 
40         return ret;     
41     }
42 };