04-树4. Root of AVL Tree (25)
04-树4. Root of AVL Tree (25)
An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illustrate the
rotation rules.


Now given a sequence of insertions, you are supposed to tell the root of the resulting AVL tree.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=20) which is the total number of keys to be inserted. Then N distinct integer keys are given in the next line. All the numbers in a line are separated by
a space.
Output Specification:
For each test case, print ythe root of the resulting AVL tree in one line.
Sample Input 1:
5
88 70 61 96 120
Sample Output 1:
70
Sample Input 2:
7
88 70 61 96 120 90 65
Sample Output 2:
88
#include <stdio.h>
struct Node {
int val;
int height;
struct Node *left;
struct Node *right;
};
int max(int a, int b) { //返回两者较大者
return a > b ? a : b;
}
int height(struct Node* root) { //为了兼容空树,树高度不能直接返回根节点的height属性
if (root == NULL) {
return -1;
}
else {
return root->height;
}
}
struct Node* RRrotation(struct Node* k1) { //右右旋转
struct Node* k2 = k1->right; //k2为根节点k1的右儿子
k1->right = k2->left; //将k2的左儿子连接到k1的右子节点
k2->left = k1; //将k1连接到k2的左子节点
k1->height = max(height(k1->left), height(k1->right)) + 1; //更新节点高度,仅仅有k1,k2节点高度变化
k2->height = max(height(k2->left), height(k2->right)) + 1;
return k2;
}
struct Node* LLrotation(struct Node* k1) { //左左旋转
struct Node* k2 = k1->left;
k1->left = k2->right;
k2->right = k1;
k1->height = max(height(k1->left), height(k1->right)) + 1;
k2->height = max(height(k2->left), height(k2->right)) + 1;
return k2;
}
struct Node* RLrotation(struct Node* k1) { //右左旋转
//分两步:先对根节点的右子树做左左旋转。再对根做右右旋转
k1->right = LLrotation(k1->right);
return RRrotation(k1);
}
struct Node* LRrotation(struct Node* k1) { //左右旋转
k1->left = RRrotation(k1->left);
return LLrotation(k1);
}
struct Node* insertAvlTree(struct Node* node, struct Node* root) {
if (root == NULL) {
root = node;
return root;
}
if (node->val > root->val) {
root->right = insertAvlTree(node, root->right); //插入右子树
if (height(root->right) - height(root->left) == 2) {
if (node->val > root->right->val) { //假设插入右子树的右子树,进行右右旋转
root = RRrotation(root);
}
else if (node->val < root->right->val) { //进行右左旋转
root = RLrotation(root);
}
}
}
else if (node->val < root->val) { //插入左子树情况与上面相似
root->left = insertAvlTree(node, root->left);
if (height(root->left) - height(root->right) == 2) {
if (node->val < root->left->val) {
root = LLrotation(root);
}
else if(node->val > root->left->val) {
root = LRrotation(root);
}
}
}
//递归中不断更新插入节点到根节点路径上全部节点的高度
root->height = max(height(root->left), height(root->right)) + 1;
return root;
}
int main() {
freopen("test.txt", "r", stdin);
int n;
scanf("%d", &n);
struct Node nodes[20];
struct Node *root = NULL;
for (int i = 0; i < n; ++i) { //初始化一个节点。并插入AVL树中
scanf("%d", &nodes[i].val);
nodes[i].height = 0; //孤立的节点高度为0
nodes[i].left = NULL;
nodes[i].right = NULL;
root = insertAvlTree(&nodes[i], root);
}
printf("%d", root->val);
return 0;
}
题目链接:http://www.patest.cn/contests/mooc-ds/04-%E6%A0%914
04-树4. Root of AVL Tree (25)的更多相关文章
- pat04-树4. Root of AVL Tree (25)
04-树4. Root of AVL Tree (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 CHEN, Yue An A ...
- pat 甲级 1066. Root of AVL Tree (25)
1066. Root of AVL Tree (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue An A ...
- PTA 04-树5 Root of AVL Tree (25分)
题目地址 https://pta.patest.cn/pta/test/16/exam/4/question/668 5-6 Root of AVL Tree (25分) An AVL tree ...
- PAT甲级:1066 Root of AVL Tree (25分)
PAT甲级:1066 Root of AVL Tree (25分) 题干 An AVL tree is a self-balancing binary search tree. In an AVL t ...
- pat1066. Root of AVL Tree (25)
1066. Root of AVL Tree (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue An A ...
- PAT 甲级 1066 Root of AVL Tree (25 分)(快速掌握平衡二叉树的旋转,内含代码和注解)***
1066 Root of AVL Tree (25 分) An AVL tree is a self-balancing binary search tree. In an AVL tree, t ...
- 1066 Root of AVL Tree (25分)(AVL树的实现)
An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...
- PAT Advanced 1066 Root of AVL Tree (25) [平衡⼆叉树(AVL树)]
题目 An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child ...
- PAT 1066. Root of AVL Tree (25)
An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...
随机推荐
- python笔记11-多线程之Condition(条件变量)
前言 当小伙伴a在往火锅里面添加鱼丸,这个就是生产者行为:另外一个小伙伴b在吃掉鱼丸就是消费者行为.当火锅里面鱼丸达到一定数量加满后b才能吃,这就是一种条件判断了. 这就是本篇要讲的Condition ...
- MySQL 5.7 安装完成后,首次登陆的几个问题
Server:CentOS 7.0 MySQL : 5.7.20 MySQL Community Server (GPL) 1.首次登陆后修改密码: 根据安装时的选择不同,有mysqld_safe用m ...
- Unity 游戏开发技巧集锦之使用cookie类型的纹理模拟云层的移动
Unity 游戏开发技巧集锦之使用cookie类型的纹理模拟云层的移动 使用cookie类型的纹理模拟云层的移动 现实生活中,当阳光直射大地,而天空中又有很多云时,云层的影子总是会投射在大地上,风吹着 ...
- POJ2955【区间DP】
题目链接[http://poj.org/problem?id=2955] 题意:[].()的匹配问题,问一个[]()串中匹配的字符数,匹配方式为[X],(X),X为一个串,问一个长度为N(N<= ...
- SpringMVC 常用注解 详解
SpringMVC 常用注解 详解 SpringMVC 常用注解 1.@RequestMapping 路径映射 2.@Requ ...
- ZOJ 1015 弦图判定
一些定义: 弦图是一种特殊图:它的所有极小环都只有3个顶点. 单纯点:该顶点与其邻接点在原图中的导出子图是一个完全图. 图G的完美消去序列:一个顶点序列a1a2a3...an,使得对于每个元素ai,a ...
- JSON 与 JS 对象的关系
很多人搞不清楚 JSON 和 Js 对象的关系,甚至连谁是谁都不清楚.简单来说: JSON 是 JS 对象的字符串表示法,它使用文本表示一个 JS 对象的信息,本质是一个字符串. 如 var obj ...
- Codeforces Round #305 (Div. 1) B. Mike and Feet 单调栈
B. Mike and Feet Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/547/pro ...
- java 随机数种子
引子:需要实现每天随机获得一个礼包,且全服玩家随出来的都是同一个. 实现方案:以当前时间是一年的第几天作为random的种子,取1~礼包总个数范围内的随机值. public static int ge ...
- Spring Boot 中文参考文档
本人翻译Spring Boot官方文档已经有几天了,虽然不能全天投入,但是间隔一到两天时间还是可以翻译部分内容,Spring Boot的文档内容比较多,翻译工作注定是长期的,毕竟个人能力有限,但为了加 ...