

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
unordered_map<int,int> in_map;
int res = 0;
int pathSum(TreeNode* root, int targetSum) {
if(root==NULL)
return 0;
in_map[0] = 1; // 前缀和为0的路径只有一条:哪个节点都不选 这个必须加上
presum(root, 0 , targetSum);
return res;
}
void presum(TreeNode* root, int cursum, int targetSum){
if(root==NULL)
return;
cursum = cursum + root->val;
if(in_map.find(cursum-targetSum)!=in_map.end()){
res = res + in_map[cursum-targetSum];
// 当前路径中存在以当前节点为终点的和为sum的子路径
}
in_map[cursum]++; // 将当前节点加入路径
presum(root->left, cursum, targetSum);
presum(root->right, cursum, targetSum);
in_map[cursum]--; // 回溯
}
};