Binary Tree Level Order Traversal II

import java.util.LinkedList;import java.util.List;import java.util.Queue;/* * 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,null,null,15,7],    3   / \  9  20    /  \   15   7return its bottom-up level order traversal as:[  [15,7],  [9,20],  [3]] * */public class Solution {public static void main(String[] args) {// TODO Auto-generated method stub}//采用DFS深度遍历public List<List<Integer>> levelOrderBottom(TreeNode root) {        Queue<TreeNode> que = new LinkedList<>();        List<List<Integer>> list = new LinkedList<List<Integer>>();        if(root == null)        return list;        que.offer(root);//先将根压入队列        while(!que.isEmpty()){//若队列不为空        int nodenums = que.size();//这一层共有多少个节点        List<Integer> subli = new LinkedList<>();        for(int i = 0;i<nodenums;i++)//对该层的每个节点进行遍历        {        if(que.peek().left!=null)//去除队列尾部节点，判断左右节点是否为空，不为空则压入队列，等到下次遍历        que.offer(que.peek().left);        if(que.peek().right!=null)        que.offer(que.peek().right);            subli.add(que.poll().val);//将该节点值加入到链表中        }        list.add(0,subli);//将该层值以链表形式加入到外层链表中，并且按要求加入到链表的开头        }        return list;    }}class TreeNode {    int val;    TreeNode left;    TreeNode right;    TreeNode(int x)     {     val = x;    }}