437.path-sum-iii.cpp

//  Tag: Tree, Depth-First Search, Binary Tree
//  Time: O(N^2)
//  Space: O(H)
//  Ref: -
//  Note: -

//  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:
//  
//  The number of nodes in the tree is in the range [0, 1000].
//  -109 <= Node.val <= 109
//  -1000 <= targetSum <= 1000
//  
//  

/**
 * 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:
    int pathSum(TreeNode* root, int targetSum) {
        if (!root) {
            return 0;
        }

        return nodeSum(root, targetSum) + pathSum(root->left, targetSum) + pathSum(root->right, targetSum);
    }

    int nodeSum(TreeNode* root, long targetSum) {
        if (!root){
            return 0;
        }

        int count = root->val == targetSum ? 1: 0;
        count+= nodeSum(root->left, targetSum - root->val);
        count+= nodeSum(root->right, targetSum - root->val);
        return count;
    }
};