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:

5 -> 3

5 -> 2 -> 1

-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);
}
};