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;
     }
 }