04_Path_sum.cpp

// Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

// A leaf is a node with no children.

// Example 1:

// Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
// Output: true
// Explanation: The root-to-leaf path with the target sum is shown.
// Example 2:

// Input: root = [1,2,3], targetSum = 5
// Output: false
// Explanation: There are two root-to-leaf paths in the tree:
// (1 --> 2): The sum is 3.
// (1 --> 3): The sum is 4.
// There is no root-to-leaf path with sum = 5.
// Example 3:

// Input: root = [], targetSum = 0
// Output: false
// Explanation: Since the tree is empty, there are no root-to-leaf paths.








/**
 * 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:
    bool hasPathSum(TreeNode* root, int targetSum) {
        function<bool(TreeNode*, int)> dfs = [&](TreeNode* node, int currentSum) -> bool    {
            if (!node) {
                return false;
            }
            currentSum += node->val;
            if (!node->left && !node->right && currentSum == targetSum) {
                return true;
            }
            return dfs(node->left, currentSum) || dfs(node->right, currentSum);
        };
        return dfs(root, 0);
    }
};