牛客题霸NC14java版答案
作者:互联网
题目描述
给定一个二叉树,返回该二叉树的之字形层序遍历,(第一层从左向右,下一层从右向左,一直这样交替)
例如:
给定的二叉树是{3,9,20,#,#,15,7},
题解
本题具体代码如下:
import java.util.*;
/*
* public class TreeNode {
* int val = 0;
* TreeNode left = null;
* TreeNode right = null;
* }
*/
public class Solution {
/**
*
* @param root TreeNode类
* @return int整型ArrayList<ArrayList<>>
*/
public ArrayList<ArrayList<Integer>> zigzagLevelOrder (TreeNode root) {
ArrayList<ArrayList<Integer>> res = new ArrayList();
if(root == null) return res;
Queue<TreeNode> queue = new LinkedList();
queue.offer(root);
boolean flag = false; //用于判断是否需要反转
while(!queue.isEmpty()){
ArrayList<Integer> tem = new ArrayList();
int size = queue.size(); //当前队列中元素得个数
for(int i = 0; i < size; i++){
TreeNode cur = queue.poll();
tem.add(cur.val);
if(cur.left != null) queue.offer(cur.left);
if(cur.right != null) queue.offer(cur.right);
}
if(!flag){
res.add(tem);
flag = !flag;
}
else{
Collections.reverse(tem);
res.add(tem);
flag = !flag;
}
}
return res;
}
}
二叉树层序遍历
import java.util.*;
/*
* public class TreeNode {
* int val = 0;
* TreeNode left = null;
* TreeNode right = null;
* }
*/
public class Solution {
/**
*
* @param root TreeNode类
* @return int整型ArrayList<ArrayList<>>
*/
public ArrayList<ArrayList<Integer>> zigzagLevelOrder (TreeNode root) {
ArrayList<ArrayList<Integer>> res = new ArrayList();
if(root == null) return res;
Queue<TreeNode> queue = new LinkedList();
queue.offer(root);
while(!queue.isEmpty()){
ArrayList<Integer> tem = new ArrayList();
int size = queue.size();
for(int i = 0; i < size; i++){
TreeNode cur = queue.poll();
tem.add(cur.val);
if(cur.left != null) queue.offer(cur.left);
if(cur.right != null) queue.offer(cur.right);
}
res.add(tem);
}
return res;
}
}
代码解释:
本题属于二叉树层序遍历得变形,利用队列暂存每一层得元素,将每一层遍历得结果存入结果中,由于是之字形遍历,所以用flag作为一个标记,用于判断是否需要反转。
标签:TreeNode,cur,res,ArrayList,queue,答案,null,客题,NC14java 来源: https://www.cnblogs.com/fangyenan/p/13959317.html