Easy
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.
Constraints:
[0, 5000].-1000 <= Node.val <= 1000-1000 <= targetSum <= 1000import com_github_leetcode.TreeNode;
/*
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
if (sum == root.val && root.left == null && root.right == null) {
return true;
}
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}