pathSum.js

/*
112. Path Sum
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 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.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {boolean}
 */
var hasPathSum = function (root, targetSum) {
    let isPathSum = false

    const traverse = (root, sum) => {
        if (root == null) return false

        if (!root.left && !root.right) return sum == root.val

        const left = traverse(root.left, sum - root.val)
        const right = traverse(root.right, sum - root.val)

        return left || right
    }

    isPathSum = traverse(root, targetSum)

    return isPathSum
};

// Better
var hasPathSum = function (root, targetSum) {
    if (!root) return false

    if (!root.left && !root.right && targetSum - root.val === 0) return true

    const isLeftTreeHasPathSum = hasPathSum(root.left, targetSum - root.val)
    const isRightTreeHasPathSum = hasPathSum(root.right, targetSum - root.val)

    return isLeftTreeHasPathSum || isRightTreeHasPathSum
};