Balanced Binary Tree Check

/*
 * Problem : Process a binary tree.
 * Input   : One line: the tree in level order, null for a missing child.
 * Output  : Balanced or not.
 * Approach: Height with a -1 mark.
 * Time    : O(n)
 * Space   : O(h)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
        // Step 1: A tree node has a value and links to the left and right child. null means: no child.
        static class TreeNode {
            int value;
            TreeNode left;
            TreeNode right;

            TreeNode(int value) {
                this.value = value;
            }
        }

        // This method builds a tree from the level order tokens. A queue keeps the nodes that still need children.
        static TreeNode build(String[] tokens) {
            if (tokens.length == 0 || tokens[0].equals("null")) {
                return null;
            }
            TreeNode root = new TreeNode(Integer.parseInt(tokens[0]));
            java.util.ArrayDeque<TreeNode> queue = new java.util.ArrayDeque<>();
            queue.add(root);
            int i = 1;
            while (!queue.isEmpty() && i < tokens.length) {
                TreeNode node = queue.poll();
                if (i < tokens.length && !tokens[i].equals("null")) {
                    node.left = new TreeNode(Integer.parseInt(tokens[i]));
                    queue.add(node.left);
                }
                i++;
                if (i < tokens.length && !tokens[i].equals("null")) {
                    node.right = new TreeNode(Integer.parseInt(tokens[i]));
                    queue.add(node.right);
                }
                i++;
            }
            return root;
        }

        // Step 2: Returns the height of the tree, or -1 if some node below is not balanced.
        static int check(TreeNode node) {
            if (node == null) {
                return 0;
            }
            int left = check(node.left);
            if (left < 0) {
                return -1;
            }
            int right = check(node.right);
            if (right < 0) {
                return -1;
            }
            // Step 3: This node is balanced when the two heights differ by at most 1.
            return Math.abs(left - right) <= 1 ? 1 + Math.max(left, right) : -1;
        }

    // This method solves the task. It receives the input and returns the answer.
    static boolean solve(String s) {
        return check(build(s.trim().split(" "))) >= 0;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s = scanner.hasNextLine() ? scanner.nextLine() : "";
        System.out.println(solve(s));
    }
}