# Path Sum II Problem

## Description

LeetCode Problem 113.

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.

A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.

Example 1:

``````1
2
3
4
5
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22
``````

Example 2:

``````1
2
Input: root = [1,2,3], targetSum = 5
Output: []
``````

Example 3:

``````1
2
Input: root = [1,2], targetSum = 0
Output: []
``````

Constraints:

• The number of nodes in the tree is in the range [0, 5000].
• -1000 <= Node.val <= 1000
• -1000 <= targetSum <= 1000

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
class Solution {
public:
void getPathSum(TreeNode* root, int sum, int curr_sum, vector<int>& path, vector<vector<int>>& ans) {
if (root == NULL)
return;
curr_sum += root->val;
path.push_back(root->val);
if ((root->left == NULL) && (root->right == NULL)) {
if (curr_sum == sum) {
ans.push_back(path);
}
}
if (root->left != NULL)
getPathSum(root->left, sum, curr_sum, path, ans);
if (root->right != NULL)
getPathSum(root->right, sum, curr_sum, path, ans);
curr_sum -= root->val;
path.pop_back();
}

vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> ans;
vector<int> path;
getPathSum(root, sum, 0, path, ans);
return ans;
}
};
``````