import java.util.ArrayList;
import java.util.Arrays;
import java.util.List; /**
* Source : https://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/
*
*
* Given a singly linked list where elements are sorted in ascending order,
* convert it to a height balanced BST.
*/
public class ConvertSortedList { /**
* 将一个有序链表转化为一棵AVL树
* 可以先把单向链表转化为有序数组,然后将数组转化为AVL树
* 或者使用双指针,快指针每次移动两个位置,慢指针每次移动一个位置,这样就能找到中间的节点
*
* 还有一种方法:
* 找到链表总个数total,一次递归如下:
* 递归查找左子树,每次递归的时候将head指向下一个节点
* 左子树递归完成之后,head指向了mid节点
* 构造root节点,当前head指向的就是mid节点,也就是root节点
* 递归构造右子树
* 构造根节点,并返回
*
* @param head
* @return
*/
public TreeNode convert (ListNode head) {
ListNode node = head;
int total = 0;
while (node != null) {
total++;
node = node.next;
}
return convert(new HeadHolder(head), 0, total-1);
} /**
* 因为每次递归需要改变head的值,所以使用一个HeadHolder指向head,每次修改headHolder的指向
*
* @param head
* @param left
* @param right
* @return
*/
public TreeNode convert (HeadHolder head, int left, int right) {
if (left > right) {
return null;
}
int mid = (left + right) / 2;
TreeNode leftChild = convert(head, left, mid-1);
// 上面递归完成之后,head指向了mid位置的节点
TreeNode root = new TreeNode(head.head.value);
// head的下一个节点就是右子树的第一个节点
head.head = head.head.next;
TreeNode rightChild = convert(head, mid+1, right);
root.leftChild = leftChild;
root.rightChild = rightChild;
return root;
} private class HeadHolder {
ListNode head; public HeadHolder(ListNode head) {
this.head = head;
}
} public ListNode createList (int[] arr) {
if (arr.length == 0) {
return null;
}
ListNode head = new ListNode();
head.value = arr[0];
ListNode pointer = head;
for (int i = 1; i < arr.length; i++) {
ListNode node = new ListNode();
node.value = arr[i];
pointer.next = node;
pointer = pointer.next;
}
return head;
}
public void binarySearchTreeToArray (TreeNode root, List<Character> chs) {
if (root == null) {
chs.add('#');
return;
}
List<TreeNode> list = new ArrayList<TreeNode>();
int head = 0;
int tail = 0;
list.add(root);
chs.add((char) (root.value + '0'));
tail ++;
TreeNode temp = null; while (head < tail) {
temp = list.get(head);
if (temp.leftChild != null) {
list.add(temp.leftChild);
chs.add((char) (temp.leftChild.value + '0'));
tail ++;
} else {
chs.add('#');
}
if (temp.rightChild != null) {
list.add(temp.rightChild);
chs.add((char)(temp.rightChild.value + '0'));
tail ++;
} else {
chs.add('#');
}
head ++;
} //去除最后不必要的
for (int i = chs.size()-1; i > 0; i--) {
if (chs.get(i) != '#') {
break;
}
chs.remove(i);
}
} private class TreeNode {
TreeNode leftChild;
TreeNode rightChild;
int value; public TreeNode(int value) {
this.value = value;
} public TreeNode() { }
} private class ListNode {
ListNode next;
int value; public ListNode(int value) {
this.value = value;
} public ListNode() { }
} public static void main(String[] args) {
ConvertSortedList convertSortedList = new ConvertSortedList();
int[] arr = new int[]{1,2,3,4,5,6};
List<Character> chs = new ArrayList<Character>();
TreeNode root = convertSortedList.convert(convertSortedList.createList(arr));
convertSortedList.binarySearchTreeToArray(root, chs);
System.out.println(Arrays.toString(chs.toArray(new Character[chs.size()])));
}
}

leetcode — convert-sorted-list-to-binary-search-tree的更多相关文章

  1. LeetCode:Convert Sorted Array to Binary Search Tree,Convert Sorted List to Binary Search Tree

    LeetCode:Convert Sorted Array to Binary Search Tree Given an array where elements are sorted in asce ...

  2. Leetcode: Convert sorted list to binary search tree (No. 109)

    Sept. 22, 2015 学一道算法题, 经常回顾一下. 第二次重温, 决定增加一些图片, 帮助自己记忆. 在网上找他人的资料, 不如自己动手. 把从底向上树的算法搞通俗一些. 先做一个例子: 9 ...

  3. [LeetCode] Convert Sorted List to Binary Search Tree 将有序链表转为二叉搜索树

    Given a singly linked list where elements are sorted in ascending order, convert it to a height bala ...

  4. [LeetCode] Convert Sorted Array to Binary Search Tree 将有序数组转为二叉搜索树

    Given an array where elements are sorted in ascending order, convert it to a height balanced BST. 这道 ...

  5. leetcode -- Convert Sorted List to Binary Search Tree

    Given a singly linked list where elements are sorted in ascending order, convert it to a height bala ...

  6. LeetCode: Convert Sorted List to Binary Search Tree 解题报告

    Convert Sorted List to Binary Search Tree Given a singly linked list where elements are sorted in as ...

  7. LeetCode: Convert Sorted Array to Binary Search Tree 解题报告

    Convert Sorted Array to Binary Search Tree Given an array where elements are sorted in ascending ord ...

  8. LeetCode——Convert Sorted List to Binary Search Tree

    Given a singly linked list where elements are sorted in ascending order, convert it to a height bala ...

  9. [LeetCode] Convert Sorted List to Binary Search Tree DFS,深度搜索

    Given a singly linked list where elements are sorted in ascending order, convert it to a height bala ...

  10. LeetCode——Convert Sorted Array to Binary Search Tree

    Description: Given an array where elements are sorted in ascending order, convert it to a height bal ...

随机推荐

  1. Codeforces.1129E.Legendary Tree(交互 二分)

    题目链接 \(Description\) 有一棵\(n\)个点的树.你需要在\(11111\)次询问内确定出这棵树的形态.每次询问你给定两个非空且不相交的点集\(S,T\)和一个点\(u\),交互库会 ...

  2. 2018-2019-2 网络对抗技术 20162329 Exp2 后门原理与实践

    目录 1.实践基础 1.1.什么是后门 1.2.基础问题 2.实践内容 2.1.使用netcat获取主机操作Shell,cron启动 2.2.使用socat获取主机操作Shell, 任务计划启动 2. ...

  3. Rectangular Covering [POJ2836] [状压DP]

    题意 平面上有 n (2 ≤ n ≤ 15) 个点,现用平行于坐标轴的矩形去覆盖所有点,每个矩形至少盖两个点,矩形面积不可为0,求这些矩形的最小面积. Input The input consists ...

  4. ionic-基于angularjs实现的多级城市选择组件

    大家都知道在移动端的选择地区组件,大部分都是模拟IOS选择器做的城市三级联动,但是在IOS上比较好,在Android上因为有的不支持ion-scroll.所以就会出现滚动不会自动回滚到某一个的正中间. ...

  5. Java语法细节 - 内存和枚举

    目录 Java申请DirectBuffer ByteBuffer的position,limit,capacity,flip操作之间的关系 枚举实现单例模式 Java申请DirectBuffer /*- ...

  6. redis离线集群安装

    用一个叫redis-trib.rb的ruby脚本.redis-trib.rb是redis官方推出的管理redis集群的工具,集成在redis的源码src目录下(redis-xxx/src/).是基于r ...

  7. Winsock编程基础1

    1.加载和释放Winsoke库 //所有Winsock函数都是从WS2_32.DLL导出,包含相应库文件#include <winsock2.h>#pragma comment(lib, ...

  8. Mesos源码分析(16): mesos-docker-executor的运行

    mesos-docker-executor的运行代码在src/docker/executor.cpp中   int main(int argc, char** argv) {   GOOGLE_PRO ...

  9. 最小可用id

    题目:在非负数组(乱序)中找到最小的可分配的id(从1开始编号),数据量10000000. 题目解读:在一个不重复的乱序的自然数组中找到最小的缺失的那个数,比如1,2,3,6,4,5,8,11.那么最 ...

  10. 全面解密QQ红包技术方案:架构、技术实现、移动端优化、创新玩法等

    本文来自腾讯QQ技术团队工程师许灵锋.周海发的技术分享. 一.引言 自 2015 年春节以来,QQ 春节红包经历了企业红包(2015 年).刷一刷红包(2016 年)和 AR 红包(2017 年)几个 ...