You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

思路：深度优先搜索，分别以root, root->left, root->right作为开始节点，计算下方是否有满足和为sum的路径。dfs用于统计个数。

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11     void dfs(TreeNode* root, int sum, int &num) {
12         if (root == NULL)
13             return;
14         if (sum == root->val)
15             num++;
16         dfs(root->left, sum - root->val, num);
17         dfs(root->right, sum - root->val, num);
18     }
19 public:
20     int pathSum(TreeNode* root, int sum) {
21         if (root == NULL) {
22             return 0;
23         }
24         int num = 0;
25         dfs(root, sum, num);
26         int left = pathSum(root->left, sum);
27         int right = pathSum(root->right, sum);
28         return num + left + right;
29     }
30 };