> For the complete documentation index, see [llms.txt](https://imhuay.gitbook.io/studies/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://imhuay.gitbook.io/studies/algorithms/problems/2022/06/leetcode0112-jian-dan-lu-jing-zong-he.md).

# 路径总和

![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/problems/2022/06/pages/R5NyOzkn3qAZy7wCx1pS#简单) [![](https://img.shields.io/static/v1?label=\&message=LeetCode\&color=green\&style=flat-square)](/studies/algorithms.md#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/problems/2022/06/pages/R5NyOzkn3qAZy7wCx1pS#二叉树树)

**问题简述**

```
给你二叉树的根节点 root 和一个表示目标和的整数 targetSum 。判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum 。如果存在，返回 true ；否则，返回 false 。

叶子节点 是指没有子节点的节点。
```

> [112. 路径总和 - 力扣（LeetCode）](https://leetcode-cn.com/problems/path-sum/)

**思路**

* 先序遍历，达到叶子节点是进行判断；
* 注意空节点的判断；

<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 hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:

        def dfs(x, rest):
            if not x:
                return False
            
            rest -= x.val
            if not x.left and not x.right:
                return rest == 0
            l, r = dfs(x.left, rest), dfs(x.right, rest)
            rest += x.val
            return l or r
        
        ret = dfs(root, targetSum)
        return ret
```

</details>