leetcode binary-tree-level-order-traversal-ii倒序输出树的每一层

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],
]

解法一：
每次将list保存到结果list的0下标的位置

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        LinkedList<TreeNode> que = new LinkedList<TreeNode>();
        if(root==null) return res;
        que.add(root);
        while(!que.isEmpty()){
            int size = que.size();
            ArrayList<Integer> list = new ArrayList<Integer>();
            for(int i=0;i<size;i++){
                TreeNode cur = que.pop();
                list.add(cur.val);
                if(cur.left!=null){
                    que.add(cur.left);
                }
                if(cur.right!=null){
                    que.add(cur.right);
                }
            }
            res.add(0,list);
        }
        return res;
    }
}

解法二：
用递归实现层序遍历 与正常遍历不同的是，
先进行下一层递归，再把当前层的结果保存到res中

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    private ArrayList<ArrayList<Integer>> res;
    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
        res = new ArrayList<ArrayList<Integer>>();
        LinkedList<TreeNode> que = new LinkedList<TreeNode>();
        if(root==null) return res;
        que.add(root);
        gao(que);
        return res;
    }
    public void gao(LinkedList<TreeNode> que){
        int size = que.size();
        if(size==0) return ;
        ArrayList<Integer> list = new ArrayList<Integer>();
        for(int i=0;i<size;i++){
            TreeNode cur = que.pop();
            list.add(cur.val);
            if(cur.left!=null){
                que.add(cur.left);
            }
            if(cur.right!=null){
                que.add(cur.right);
            }
        }
        gao(que);
        res.add(list);
    }
}