107. Binary Tree Level Order Traversal II Leetcode Java

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]
]

Solution:

Actually, I didn't come out a very "good" / special solution for this problem, I just simply reverse the result of level order traversal. 

 public List<List<Integer>> levelOrderBottom(TreeNode root) {  
     List<List<Integer>> res=new ArrayList<List<Integer>>();  
     if(root==null) return res;  
     LinkedList<TreeNode> q=new LinkedList<TreeNode>();  
     q.offer(root);  
     int curL=0;  
     int lastL=1;  
     List<Integer> items=new ArrayList<Integer>();  
     while(!q.isEmpty()){  
      TreeNode cur=q.poll();  
      lastL--;  
      items.add(cur.val);  
      if(cur.left!=null){  
        q.offer(cur.left);  
        curL++;  
      }  
      if(cur.right!=null){  
        q.offer(cur.right);  
        curL++;  
      }  
      if(lastL==0){  
        res.add(items);  
        items=new ArrayList<Integer>();  
        lastL=curL;  
        curL=0;  
      }  
     }  
     Collections.reverse(res);  
     return res;  
   }