Binary Tree Level Order Traversal II --- LeetCode

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.

解题思路：同Binary Tree Level Order Traversal, 使用两个队列来遍历，递归遍历的同时记录每层的节点值到list中，遍历到叶节点后倒序添加到List<List<Integer>>结果集合中。

/**
 * 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>> traversal(Queue<TreeNode> queue,List<List<Integer>> rs){
       List<Integer> list=new ArrayList<Integer>();
       Queue<TreeNode> q=new LinkedList<TreeNode>();
       while(!queue.isEmpty()){
        TreeNode node=queue.poll();
        if(node!=null){
        list.add(node.val);
        if(node.left!=null){
        q.offer(node.left);
        }
        if(node.right!=null){
        q.offer(node.right);
        }
        }        
       }
       if(!q.isEmpty()){
        rs=traversal(q,rs);
       }   
        rs.add(list);
   return rs;
   }

    public List<List<Integer>> levelOrderBottom(TreeNode root) {
           List<List<Integer>> rs=new ArrayList<List<Integer>>();
           Queue<TreeNode> queue=new LinkedList<TreeNode>();
           if(root==null){
          return rs;
           }
           queue.offer(root);
           rs=traversal(queue,rs);                    
      return rs;
    }
}