# LeetCode –Binary Tree Preorder Traversal
Date: 2020-04-16


### **题目：**

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

**Example:**

**Input:**

```
 [1,null,2,3]
   1
    \
     2
    /
   3

```

**Output:**

```
 [1,3,2]
```

**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，然后移动到左节点， 保证“根-左”顺序， 为空后出栈，取其右节点入栈中。这样就保证了访问顺序为“根-左-右”。

```
/**
 * 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> preorderTraversal(TreeNode* root) {
        
        vector<int> res;
        stack<TreeNode*> ss;
        TreeNode *p = root;
        while (!ss.empty()||p){
            if (p!=NULL){
                ss.push(p);
                res.push_back(p->val);
                p = p->left;
            }
            else{
                p = ss.top();
                ss.pop();
                p=p->right;
                
            }
        }
        return res;
        
    }
};
```

### **实现：**

- 递归实现

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

### 参考:

[\[LeetCode\] 144. Binary Tree Preorder Traversal 二叉树的先序遍历](https://www.cnblogs.com/grandyang/p/4146981.html)

[Tree Preorder Traversal 3 different solutions](https://leetcode.com/problems/binary-tree-preorder-traversal/discuss/45468/3-Different-Solutions)