# Path Sum II

> Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

> For example:
Given the below binary tree and sum = 22.


```
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1

```

> return

```
[
[5,4,11,2],
[5,8,4,5]
]
```
题目翻译:
给定一个二叉树,并且给定一个值,找出所有从根节点到叶子节点和等于这个给定值的路径.上面的例子可以很好地让读者理解这个题目的目的.

解题思路:
这个题目和Path Sum的解法几乎是一模一样,都是用dfs来进行求解,不过就是在传参数的时候有些不同了,因为题目的要求也不同.

时间复杂度:
O(n)

代码如下:

```c++

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector > pathSum(TreeNode *root, int sum) {
vector> ret;
if(root == NULL)
return ret;
vector curr;
DFS(ret,curr,sum,0,root);
return ret;
}

void DFS(vector>& ret, vector curr, int sum, int tmpsum, TreeNode* root)
{
if(root == NULL)
return;
tmpsum+=root->val;
curr.push_back(root->val);
if(tmpsum == sum)
{
if(root->left == NULL&&root->right == NULL)
{
ret.push_back(curr);
return;
}
}
DFS(ret,curr,sum,tmpsum,root->left);
DFS(ret,curr,sum,tmpsum,root->right);
}
};
```