Path Sum III

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
   方法： DFS

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {private:    int find(TreeNode* root,int temp_sum,int& sum){        if(root==NULL)            return 0;        int temp = temp_sum + root->val;        return (temp==sum) + find(root->left,temp,sum) + find(root->right,temp,sum);    }public:    int pathSum(TreeNode* root, int sum) {        if(root == NULL)            return 0;        return find(root,0,sum) + pathSum(root->left,sum) + pathSum(root->right,sum);    }};