# Path Sum

> Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

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

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

```
> return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

题目翻译:
给定一颗二叉树和一个特定值,写一个方法来判定这棵树是否存在这样一种条件,使得从root到其中一个叶子节点的路径的和等于给定的sum值.

解题思路:
这道题很常规,直接用DFS就可以求解.

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:
bool hasPathSum(TreeNode *root, int sum) {
if(root == NULL)
return false;
return DFS(sum, 0, root);
}

bool DFS(int target, int sum, TreeNode* root)
{
if(root == NULL)
return false;
sum += root->val;
if(root->left == NULL && root->right == NULL)
{
if(sum == target)
return true;
else
return false;
}
bool leftPart = DFS(target, sum, root->left);
bool rightPart = DFS(target, sum, root->right);
return leftPart||rightPart;
}

};

```