Binary Tree Level Order Traversal Show result

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

Have you met this question in a real interview? 

Yes

Example

Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

Challenge

Using only 1 queue to implement it.

public class Solution {

    /**

     * @param root: The root of binary tree.

     * @return: Level order a list of lists of integer

     */

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

        // write your code here

        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();

        if(root == null) return res;

        Queue<TreeNode> queue = new LinkedList<TreeNode>();

        queue.offer(root);

        while(!queue.isEmpty()){

            Queue<TreeNode> nextQueue = new LinkedList<TreeNode>();

            ArrayList<Integer> sub = new ArrayList<Integer>();

            while(!queue.isEmpty()){

                TreeNode node = queue.poll();

                sub.add(node.val);

                if(node.left != null) nextQueue.add(node.left);

                if(node.right != null) nextQueue.add(node.right);

            }

            res.add(sub);

            queue = nextQueue;

        }

        

        return res;

    }

}

public class Solution {

    /**

     * @param root: The root of binary tree.

     * @return: Level order a list of lists of integer

     */

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

        // write your code here

        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();

        if(root == null) return res;

        Queue<TreeNode> queue = new LinkedList<TreeNode>();

        queue.offer(root);

        while(!queue.isEmpty()){

            int size = queue.size();

            ArrayList<Integer> sub = new ArrayList<Integer>();

            for(int i = 0; i < size; i++){

                TreeNode node = queue.poll();

                sub.add(node.val);

                if(node.left != null) queue.add(node.left);

                if(node.right != null) queue.add(node.right);

            }

            res.add(sub);

        }

        

        return res;

    }

}