Closest Binary Search Tree Value

Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.

Note:

Given target value is a floating point.
You are guaranteed to have only one unique value in the BST that is closest to the target.

Example:

Input: root = [4,2,5,1,3], target = 3.714286

    4
   / \
  2   5
 / \
1   3

Output: 4

Solution:

/**
 * 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 int closestValue(TreeNode root, double target) {
        int[] close = new int[]{root.val};
        helper(root, target, close);
        return close[0];
    }
    
    private void helper(TreeNode root, double target, int[] close) {
        if (root == null) return;
        if (Math.abs(target - root.val) < Math.abs(target - close[0])) {
            close[0] = root.val;
        }
        if (root.val > target) {
            helper(root.left, target, close);
        } else {
            helper(root.right, target, close);
        }
    }
}