112. Path Sum

 Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

My answer is:

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        return helper(root, 0, sum);
    }
    
    private boolean helper(TreeNode node, int currentSum, int sum) {
        if (node == null) {
            return false;
        }
        currentSum += node.val;
        if (node.left == null && node.right == null) {
            return currentSum == sum;
        }
        boolean leftResult = helper(node.left, currentSum, sum);
        if (leftResult) {
            return true;
        }
        boolean rightResult = helper(node.right, currentSum, sum);
        return rightResult;
    }
}