pathSum.js

/*
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: 5 → 4 → 11 → 2

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

Example 3:
Input: root = [], targetSum = 0
Output: false

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

/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {boolean}
 */
function hasPathSum(root, targetSum) {
    if (!root) return false;
    if (!root.left && !root.right) return root.val === targetSum;
    return hasPathSum(root.left, targetSum - root.val) ||
           hasPathSum(root.right, targetSum - root.val);
}

// Helper
function TreeNode(val, left, right) { this.val = val; this.left = left || null; this.right = right || null; }

// Example usage:
const root = new TreeNode(5,
    new TreeNode(4, new TreeNode(11, new TreeNode(7), new TreeNode(2)), null),
    new TreeNode(8, new TreeNode(13), new TreeNode(4, null, new TreeNode(1)))
);
console.log(hasPathSum(root, 22)); // Output: true
console.log(hasPathSum(root, 5));  // Output: false