Medium
Given the root
of a binary tree and an integer targetSum
, return the number of paths where the sum of the values along the path equals targetSum
.
The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
Example 1:
Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
Output: 3
Explanation: The paths that sum to 8 are shown.
Example 2:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: 3
Constraints:
[0, 1000]
.-109 <= Node.val <= 109
-1000 <= targetSum <= 1000
/**
* Definition for a binary tree node.
* class TreeNode {
* int val;
* TreeNode? left;
* TreeNode? right;
* TreeNode([this.val = 0, this.left, this.right]);
* }
*/
class Solution {
int count = 0;
int pathSum(TreeNode? root, int targetSum) {
if (root == null) {
return 0;
}
helper(root, targetSum, 0);
pathSum(root.left, targetSum);
pathSum(root.right, targetSum);
return count;
}
void helper(TreeNode? node, int targetSum, int currSum) {
if (node == null) {
return;
}
currSum += node.val;
if (targetSum == currSum) {
count++;
}
helper(node.left, targetSum, currSum);
helper(node.right, targetSum, currSum);
}
}