Binary Trees2-Level order traversal

For a given a Binary Tree of type integer, print it in a level order fashion where each level will be printed on a new line. Elements on every level will be printed in a linear fashion and a single space will separate them.
Example:
alt txt

For the above-depicted tree, when printed in a level order fashion, the output would look like:

10
20 30 
40 50 60
Where each new line denotes the level in the tree.
Input Format:
The first and the only line of input will contain the node data, all separated by a single space. Since -1 is used as an indication whether the left or right node data exist for root, it will not be a part of the node data.
Output Format:
The given input tree will be printed in a level order fashion where each level will be printed on a new line. 
Elements on every level will be printed in a linear fashion. A single space will separate them.
Constraints:
1 <= N <= 10^5
Where N is the total number of nodes in the binary tree.

Time Limit: 1 sec
Sample Input 1:
10 20 30 40 50 -1 60 -1 -1 -1 -1 -1 -1 
Sample Output 1:
10 
20 30 
40 50 60 
Sample Input 2:
8 3 10 1 6 -1 14 -1 -1 4 7 13 -1 -1 -1 -1 -1 -1 -1
Sample Output 2:
8 
3 10 
1 6 14 
4 7 13 

*********************************************Solution*****************************************
import java.util.*;

public static void printLevelWise(BinaryTreeNode<Integer> root) {
		
		if (root == null)
			return;

		Queue<BinaryTreeNode<Integer>> node = new LinkedList<BinaryTreeNode<Integer>>();
		node.add(root);
		node.add(null);
		while (!node.isEmpty()) 
		{
			BinaryTreeNode<Integer> front = node.poll();
			if (front == null) 
			{
				if (node.isEmpty())
					break;
				else {
					System.out.println();
					node.add(null);
				}

			}
			else{
				System.out.print(front.data + " ");
				if (front.left != null)
					node.add(front.left);
				if (front.right != null)
					node.add(front.right);
			}
		}
	}