Binary Tree Level Order Traversal II leetcode java

题目：

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]
]

题解：
这道题跟前面一道是一样的。。
只是存到res的结果顺序不一样罢了。
之前那个就是循序的存
这道题就是每得到一个行结果就存在res的0位置，这样自然就倒序了。

代码如下：

 1     public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
 2         ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();  
 3         if(root == null)  
 4             return res;
 5         
 6         LinkedList<TreeNode> queue = new LinkedList<TreeNode>();  
 7         queue.add(root);
 8         
 9         int curLevCnt = 1;  
10         int nextLevCnt = 0;  
11         
12         ArrayList<Integer> levelres = new ArrayList<Integer>();  
13        
14         while(!queue.isEmpty()){  
15             TreeNode cur = queue.poll();  
16             curLevCnt--;  
17             levelres.add(cur.val);  
18             
19             if(cur.left != null){  
20                 queue.add(cur.left);  
21                 nextLevCnt++;  
22             }  
23             if(cur.right != null){  
24                 queue.add(cur.right);  
25                 nextLevCnt++;  
26             }  
27               
28             if(curLevCnt == 0){  
29                 curLevCnt = nextLevCnt;  
30                 nextLevCnt = 0;  
31                 res.add(0,levelres);  //insert one by one from the beginning
32                 levelres = new ArrayList<Integer>();  
33             }  
34         }  
35         return res;  
36     }