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]]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        LinkedList<TreeNode> stack=new LinkedList<TreeNode>();        LinkedList<TreeNode> curr=new LinkedList<TreeNode>();        List<List<Integer>> res=new ArrayList< List<Integer> >();        if(root==null) return res;        stack.add(root);                while(!stack.isEmpty()){            curr=stack;            stack=new LinkedList<TreeNode>();            List<Integer> a=new LinkedList<Integer>();            while(curr.size()>0){                TreeNode tmpNode=curr.peekFirst();                if(tmpNode.left!=null) stack.add(tmpNode.left);                if(tmpNode.right!=null) stack.add(tmpNode.right);                a.add(tmpNode.val);                curr.poll();            }            res.add(a);        }        return res;    }}