# 路径总和 II

<p>给你二叉树的根节点 <code>root</code> 和一个整数目标和 <code>targetSum</code> ，找出所有 <strong>从根节点到叶子节点</strong> 路径总和等于给定目标和的路径。</p>

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

<div class="original__bRMd">
<div>
<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/abdc6911d9d9ee1b12dcdbb1945a1ab2.jpeg" style="width: 500px; height: 356px;" />
<pre>
<strong>输入：</strong>root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
<strong>输出：</strong>[[5,4,11,2],[5,8,4,5]]
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/9bc801134935f549d425954a120f1cc3.jpeg" style="width: 212px; height: 181px;" />
<pre>
<strong>输入：</strong>root = [1,2,3], targetSum = 5
<strong>输出：</strong>[]
</pre>

<p><strong>示例 3：</strong></p>

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

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>树中节点总数在范围 <code>[0, 5000]</code> 内</li>
	<li><code>-1000 <= Node.val <= 1000</code></li>
	<li><code>-1000 <= targetSum <= 1000</code></li>
</ul>
</div>
</div>


## 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:
    vector<vector<int>> pathSum(TreeNode *root, int sum)
    {
        vector<vector<int>> res;
        vector<int> track;
        backTrack(root, res, track, sum);
        return res;
    }

    void backTrack(TreeNode *root, vector<vector<int>> &res, vector<int> track, int sum)
    {

        if (!root)
        {
            return;
        }
        if (!root->left && !root->right)
        {
            sum -= root->val;
            track.push_back(root->val);
            if (sum == 0)
            {
                res.push_back(track);
            }
            track.pop_back();
            sum += root->val;
            return;
        }

        sum -= root->val;
        track.push_back(root->val);

        backTrack(root->left, res, track, sum);
        backTrack(root->right, res, track, sum);

        track.pop_back();
        sum += root->val;
    }
};


```

## 答案

```cpp

```

## 选项

### A

```cpp

```

### B

```cpp

```

### C

```cpp

```