# 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://img-service.csdnimg.cn/img_convert/ef95371d9da7672aade4dc505260cc30.jpeg" 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://img-service.csdnimg.cn/img_convert/9bc801134935f549d425954a120f1cc3.jpeg" />
<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

```cpp

#include <bits/stdc++.h>
using namespace std;

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution
{
public:
    bool hasPathSum(TreeNode *root, int sum)
    {
        bool flag = false;
        backTrack(root, sum, flag);
        return flag;
    }

    void backTrack(TreeNode *root, int sum, bool &flag)
    {
        if (!root)
        {
            return;
        }

        if (!root->left && !root->right)
        {
            sum -= root->val;
            if (sum == 0)
            {
                flag = true;
            }
            sum += root->val;
            return;
        }

        sum -= root->val;
        backTrack(root->left, sum, flag);
        backTrack(root->right, sum, flag);
        sum += root->val;
    }
};
```

## 答案

```cpp

```

## 选项

### A

```cpp

```

### B

```cpp

```

### C

```cpp

```