# 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

```java

public class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;

	TreeNode() {
	}

	TreeNode(int val) {
		this.val = val;
	}

	TreeNode(int val, TreeNode left, TreeNode right) {
		this.val = val;
		this.left = left;
		this.right = right;
	}
}

class Solution {
	public boolean hasPathSum(TreeNode root, int targetSum) {
		if (root == null)
			return false;
		return traversal(root, targetSum - root.val);
	}

	public boolean traversal(TreeNode root, int count) {
		if (root.left == null && root.right == null && count == 0) {
			return true;
		}
		if (root.left == null && root.right == null) {
			return false;
		}
		if (root.left != null) {
			count -= root.left.val;
			if (traversal(root.left, count))
				return true;
			count += root.left.val;
		}
		if (root.right != null) {
			count -= root.right.val;
			if (traversal(root.right, count))
				return true;
			count += root.right.val;
		}
		return false;
	}
}
```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```