Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return [
[5,4,11,2],
[5,8,4,5]
]

  

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void DFS(TreeNode *root, vector<int> &temp, int tempSum)
{
if(root->left == NULL && root -> right== NULL && tempSum == sum) {
result.push_back(temp);
return;
} if(root->left!= NULL ){
temp.push_back(root->left->val);
DFS(root->left, temp,root->left->val + tempSum );
temp.pop_back();
} if(root->right != NULL ){
temp.push_back(root->right->val);
DFS(root->right, temp, root->right->val + tempSum);
temp.pop_back();
}
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
result.clear();
if(!root) return result;
this->sum = sum;
vector<int> temp;
temp.push_back(root->val);
DFS(root, temp, root->val);
return result;
}
private:
int sum;
vector<vector<int>> result;
};

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