# C++ Solution, one pass, very easy to understand with explanation

• The idea is simple.
For every node S, keep a vector to record all sum start from node S

``````/**
* 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:

vector<int> solver(TreeNode * root, int sum, int & count){
if(root==NULL) return {};
vector<int> left = solver(root->left,sum,count);
vector<int> right = solver(root->right,sum,count);
vector<int> res;
res.push_back(root->val);
if(root->val==sum) count++;
for(int i = 0; i < left.size(); i++){
int num = left[i] + root->val;
if(num == sum) count++;
res.push_back(num);
}
for(int i = 0; i < right.size(); i++){
int num = right[i] + root->val;
if(num == sum) count++;
res.push_back(num);
}
return res;
}

int pathSum(TreeNode* root, int sum) {
if(root==NULL) return 0;
int count = 0;
solver(root,sum,count);
return count;
}
};
``````

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