/**
* 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 {
public:
int pathSum(TreeNode* root, int sum) {
if (root == NULL)
return 0;
return Sum(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
}
private:
//pre为前面节点的和，cur为前面加上现在遍历到的节点；
int Sum(TreeNode* root, int pre, int sum){
if (root == NULL)
return 0;
int cur = pre + root->val;
return (cur == sum) + Sum(root->left, cur, sum) + Sum(root->right, cur, sum);
}
};
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
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.
