No wonder that recursion is a super fit for solving question of binary tree. Simply setting the base case of true and false, the rest is trivial.
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/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function hasPathSum(root: TreeNode | null, targetSum: number): boolean {
if ( root === null ) return false
if ( root.left === null && root.right === null && targetSum === root.val) return true
return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val)
};