[Leetcode] 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]
]

Solution:

在I的基础上倒过来就好。

/**
 * 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>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> ret=new ArrayList<List<Integer>>();
        if(root==null)
            return ret;
        List<List<Integer>> alal=new ArrayList<List<Integer>>();
        List<Integer> al=new ArrayList<Integer>();
        Queue<TreeNode> queue=new LinkedList<TreeNode>();
        queue.add(root);
        int curLevel=1;
        int nextLevel=0;
        
        while(!queue.isEmpty()){
            TreeNode cur=queue.remove();
            al.add(cur.val);
            curLevel--;
            
            if(cur.left!=null){
                queue.add(cur.left);
                nextLevel++;
            }
            
            if(cur.right!=null){
                queue.add(cur.right);
                nextLevel++;
            }
            
            if(curLevel==0){
                alal.add(al);
                al=new ArrayList<Integer>();
                curLevel=nextLevel;
                nextLevel=0;
            }
        }
        for(int i=alal.size()-1;i>=0;--i){
            ret.add(alal.get(i));
        }
        
        return ret;
    }
}