Binary Tree Level Order Traversal (LeetCode Tree)

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

Idea: Breadth-first search. Use a queue to maintain the next level.

Time: O(n) Space: O(n)

Code:

 public class Solution {  
   public List<List<Integer>> levelOrder(TreeNode root) {  
     List<List<Integer>> result=new ArrayList<List<Integer>>();  
     if(root==null)  
       return result;  
     Queue<TreeNode> queue=new LinkedList<TreeNode>();  
     queue.offer(root);  
     while(!queue.isEmpty())  
     {  
       int number=queue.size();  
       List<Integer> tmp=new ArrayList<Integer>();  
       for(int i=0;i<number;i++)  
       {  
         TreeNode cur=queue.poll();  
         tmp.add(cur.val);  
         if(cur.left!=null)  
           queue.offer(cur.left);  
         if(cur.right!=null)  
           queue.offer(cur.right);  
       }  
       result.add(tmp);  
     }  
     return result;  
   }  
 }