/**
 * 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
        
    }
};