# 二叉树的所有路径

![last modify](https://img.shields.io/static/v1?label=last%20modify\&message=2022-10-14%2014%3A59%3A33\&color=yellowgreen\&style=flat-square) [![](https://img.shields.io/static/v1?label=\&message=%E7%AE%80%E5%8D%95\&color=yellow\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/..#简单) [![](https://img.shields.io/static/v1?label=\&message=LeetCode\&color=green\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/..#leetcode) [![](https://img.shields.io/static/v1?label=\&message=%E4%BA%8C%E5%8F%89%E6%A0%91/%E6%A0%91\&color=blue\&style=flat-square)](https://imhuay.gitbook.io/studies/algorithms/..#二叉树树)

**问题简述**

```
给你一个二叉树的根节点 root ，按 任意顺序 ，返回所有从根节点到叶子节点的路径。

叶子节点 是指没有子节点的节点。
```

> [257. 二叉树的所有路径 - 力扣（LeetCode）](https://leetcode-cn.com/problems/binary-tree-paths/)

**思路**

* 先序遍历，特殊处理叶子节点；

<details>

<summary><strong>Python</strong></summary>

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        
        ret = []
        tmp = []

        def dfs(x):
            if not x: return 
            
            tmp.append(str(x.val))
            if not x.left and not x.right:
                ret.append('->'.join(tmp))
            
            dfs(x.left)
            dfs(x.right)
            tmp.pop()
        
        dfs(root)
        return ret
```

</details>