LeetCode OJ 107. Binary Tree Level Order Traversal II

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

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / 
  9  20
    /  
   15   7

return its bottom-up level order traversal as:

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

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> list = new LinkedList<List<Integer>>();
        if(root == null) return list;
        
        Queue<TreeNode> queue1 = new LinkedList<TreeNode>();
        Queue<TreeNode> queue2 = new LinkedList<TreeNode>();
        
        queue1.offer(root);
        
        while(queue1.size()>0){
            List<Integer> childlist = new ArrayList<Integer>();
            while(queue1.size()>0){
                TreeNode node = queue1.remove();
                childlist.add(node.val);
                if(node.left!=null) queue2.offer(node.left);
                if(node.right!=null) queue2.offer(node.right);
            }
            list.add(0, childlist);
            Queue<TreeNode> temp = queue1;
            queue1 = queue2;
            queue2 = temp;
        }
        
        return list;
    }
}