# 路径总和

<p>给你二叉树的根节点 <code>root</code> 和一个表示目标和的整数 <code>targetSum</code> ，判断该树中是否存在 <strong>根节点到叶子节点</strong> 的路径，这条路径上所有节点值相加等于目标和 <code>targetSum</code> 。</p>

<p><strong>叶子节点</strong> 是指没有子节点的节点。</p>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>输入：</strong>root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
<strong>输出：</strong>true
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg" />
<pre>
<strong>输入：</strong>root = [1,2,3], targetSum = 5
<strong>输出：</strong>false
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>root = [1,2], targetSum = 0
<strong>输出：</strong>false
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>树中节点的数目在范围 <code>[0, 5000]</code> 内</li>
	<li><code>-1000 <= Node.val <= 1000</code></li>
	<li><code>-1000 <= targetSum <= 1000</code></li>
</ul>


## template

```python
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution:
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if root is None:
            return False
        if sum == root.val and root.left is None and root.right is None:
            return True
        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)
```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```