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   7
return 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;
}
}