# LeetCode –Binary Tree Postorder Traversal
Date: 2020-06-04


### **题目：**

Given a binary tree, return the _postorder_ traversal of its nodes' values.

**Example:**

**Input:**

 `[1,null,2,3]` 1 \\ 2 / 3

**Output:**

 `[3,2,1]`

**Follow up:** Recursive solution is trivial, could you do it iteratively?

### **解题：**

       遍历顺序“根-左-右”

### **实现：**

- 使用栈实现

         从根节点开始，先将根节点压入栈，保存结果值，然后移动到右节点， 保证“根-右边”顺序， 为空后出栈，取其左节点入栈中。这样入栈就保证了访问顺序为“根-右-左”， 出栈实现“左-右-根”。

```
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
       vector<int> res;
       stack<TreeNode*> ss;
       TreeNode *p= root, *t;
        
       while(!ss.empty()||p){
           if (p){
               ss.push(p);
               res.insert(res.begin(), p->val);
               p = p->right;
           }
           else {
               p=ss.top();
               ss.pop();
               p= p->left;
               
           }
       }
       return res;
    }
};
```

### **实现：**

- 递归实现

```
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector <int> res;
        postorder(root, res);
        return res;
        
    }
    void postorder(TreeNode *root, vector<int> &res){
        if (!root) {
            return;
        }
        if (root->left) {
            postorder(root->left, res);
        }
        
        if (root->right){
            postorder(root->right, res);
        }
        res.push_back(root->val);
    }
};
```

### 参考:

[\[LeetCode\] 145. Binary Tree Postorder Traversal 二叉树的后序遍历](https://www.cnblogs.com/grandyang/p/4251757.html)