leetcode--Binary Tree Level Order Traversal

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]
]

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    /**This is a simple problem.<br>
	 * 
	 * @param root --the root node of a tree
	 * @return List of value of each level
	 * @author Averill Zheng
	 * @version 2016-06-03
	 * @since JDK 1.7
	 */
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer> > result = new ArrayList<List<Integer>>();
        if(root != null){
        	Queue<TreeNode> topLevel = new LinkedList<TreeNode>();
        	topLevel.add(root);
        	while(topLevel.peek() != null){
        		Queue<TreeNode> nextLevel = new LinkedList<TreeNode>();
        		List<Integer> value = new ArrayList<Integer>();
        		while(topLevel.peek() != null){
        			TreeNode node = topLevel.poll();
        			value.add(node.val);
        			if(node.left != null)
        				nextLevel.add(node.left);
        			if(node.right != null)
        				nextLevel.add(node.right);
        		}
        		result.add(value);
        		topLevel = nextLevel;
        	}
        }
        return result;    
    }
}