Level Order Traversal

/*
 * Problem : Process a binary tree.
 * Input   : One line: the tree in level order, null for a missing child.
 * Output  : One line for each level.
 * Approach: Breadth-first search by levels.
 * Time    : O(n)
 * Space   : O(w)
 * 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;
        }

    // This method solves the task. It receives the input and returns the answer.
    static String[] solve(String s) {
        TreeNode root = build(s.trim().split(" "));
        java.util.List<String> output = new java.util.ArrayList<>();
        java.util.ArrayDeque<TreeNode> queue = new java.util.ArrayDeque<>();
        queue.add(root);

        // Step 2: Process one level at a time. At the start of a round the queue holds exactly one level.
        while (!queue.isEmpty()) {
            int size = queue.size();
            StringBuilder line = new StringBuilder();

            // Step 3: Take all nodes of this level. Their children form the next level.
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (i > 0) {
                    line.append(" ");
                }
                line.append(node.value);
                if (node.left != null) {
                    queue.add(node.left);
                }
                if (node.right != null) {
                    queue.add(node.right);
                }
            }
            output.add(line.toString());
        }
        return output.toArray(new String[0]);
    }

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