Binary Tree Level Order Traversal Solutions in Java

// Source : https://oj.leetcode.com/problems/binary-tree-level-order-traversal/
// Inspired by : http://www.jiuzhang.com/solutions/binary-tree-level-order-traversal/
// Author : Lei Cao
// Date   : 2015-10-08



package binaryTreeLevelOrderTraversal;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

public class binaryTreeLevelOrderTraversal {
    /**
     * @param root: The root of binary tree.
     * @return: Level order a list of lists of integer
     */
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> results = new ArrayList<>();
        if (root == null) {
            return results;
        }

        ArrayList<Integer> values = new ArrayList<Integer>();
        Queue<TreeNode> q = new LinkedList<>();

        q.offer(root);
        q.offer(null);

        while (q.size() > 0) {
            TreeNode node = q.poll();
            // null node used as a separator of every level
            if (node == null) {
                results.add(new ArrayList<>(values));
                values.clear();
                if (q.size() == 0) {
                    break;
                }
                q.offer(null);
                continue;
            }
            values.add(node.val);
            if (node.left != null) {
                q.offer(node.left);
            }
            if (node.right != null) {
                q.offer(node.right);
            }
        }

        return results;
    }
}

/// Source : https://leetcode.com/problems/binary-tree-level-order-traversal/description/
/// Author : liuyubobobo
/// Time   : 2017-11-17

import java.util.ArrayList;
import java.util.List;
import java.util.LinkedList;
import javafx.util.Pair;

/// BFS
/// Time Complexity: O(n), where n is the number of nodes in the tree
/// Space Complexity: O(n)
class Solution {

    public List<List<Integer>> levelOrder(TreeNode root) {

        ArrayList<List<Integer>> res = new ArrayList<>();
        if(root == null)
            return res;

        LinkedList<Pair<TreeNode, Integer>> queue = new LinkedList<>();
        queue.addLast(new Pair<>(root, 0));

        while(!queue.isEmpty()){

            Pair<TreeNode, Integer> front = queue.removeFirst();
            TreeNode node = front.getKey();
            int level = front.getValue();

            if(level == res.size())
                res.add(new ArrayList<>());
            assert level < res.size();

            res.get(level).add(node.val);
            if(node.left != null)
                queue.addLast(new Pair<>(node.left, level + 1));
            if(node.right != null)
                queue.addLast(new Pair<>(node.right, level + 1));
        }

        return res;
    }
}

/// Source : https://leetcode.com/problems/binary-tree-level-order-traversal/description/
/// Author : liuyubobobo
/// Time   : 2018-10-16

import java.util.ArrayList;
import java.util.List;
import java.util.LinkedList;
import java.util.Queue;

/// BFS
/// No need to store level information in the queue :-)
///
/// Time Complexity: O(n), where n is the number of nodes in the tree
/// Space Complexity: O(n)
class Solution2 {

    public List<List<Integer>> levelOrder(TreeNode root) {

        ArrayList<List<Integer>> res = new ArrayList<>();
        if(root == null)
            return res;

        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int levelNum = 1;

        while(!queue.isEmpty()){

            int newLevelNum = 0;
            ArrayList<Integer> level = new ArrayList<>();
            for(int i = 0; i < levelNum; i ++){
                TreeNode node = queue.remove();
                level.add(node.val);

                if(node.left != null){
                    queue.add(node.left);
                    newLevelNum ++;
                }
                if(node.right != null){
                    queue.add(node.right);
                    newLevelNum ++;
                }
            }

            res.add(level);
            levelNum = newLevelNum;
        }

        return res;
    }
}