PAT甲级——1123 Is It a Complete AVL Tree (完全AVL树的判断)
嫌排版乱的话可以移步我的CSDN:https://blog.csdn.net/weixin_44385565/article/details/89390802
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 output the level-order traversal sequence of the resulting AVL tree, and to tell if it is a complete binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤ 20). 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, insert the keys one by one into an initially empty AVL tree. Then first print in a line the level-order traversal sequence of the resulting AVL tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line. Then in the next line, print YES if the tree is complete, or NO if not.
Sample Input 1:
5
88 70 61 63 65
Sample Output 1:
70 63 88 61 65
YES
Sample Input 2:
8
88 70 61 96 120 90 65 68
Sample Output 2:
88 65 96 61 70 90 120 68
NO
题目大意:给出N个数据,建立AVL树,并判断其是否为完全二叉树。
思路:题目言简意赅,就两个核心操作:建立AVL树、判断是否为完全二叉树~~
AVL树的建立过程详见我之前的文章:AVL树(自平衡二叉查找树)
因为AVL树本身的性质已经保证了左右子树的高度差≤1,所以之后判断完全二叉树主要有两个条件:1、对于每个节点,左子树高度≥右子树高度;2、层序遍历遇到一个节点,它有左孩子但没有右孩子时标记一下,在它之后进入队列的节点都为叶子节点,这棵AVL树为完全二叉树。
#include <iostream>
#include <queue>
#define ElementType int
using namespace std;
typedef struct node *AVLTree;
struct node {
ElementType key;
int Height = ;
AVLTree left = NULL, right = NULL;
};
bool flag = true;
int Height(AVLTree tree);//求树的高度
ElementType Max(ElementType a, ElementType b);
AVLTree insert(AVLTree tree, ElementType &key);//在AVLTree中插入节点
AVLTree LL_Rotation(AVLTree tree);//LL旋转
AVLTree RR_Rotation(AVLTree tree);//RR旋转
AVLTree LR_Rotation(AVLTree tree);//LR旋转
AVLTree RL_Rotation(AVLTree tree);//RL旋转
void levelTraversal(AVLTree tree);//层序遍历 int main()
{
int N;
ElementType key;
AVLTree tree = NULL;
scanf("%d", &N);
for (int i = ; i < N; i++) {
cin >> key;
tree = insert(tree, key);
}
levelTraversal(tree);
if (flag)
printf("YES\n");
else
printf("NO\n"); } AVLTree insert(AVLTree tree, ElementType &key) {
if (tree == NULL) {
tree = new node();
tree->key = key;
}
else if (key < tree->key) {
tree->left = insert(tree->left, key);//key小于当前节点的值时继续往其左子树递归地插入
if (Height(tree->left) - Height(tree->right) >= ) {//左子树与右子树的高度差达到2的时候就要对当前节点进行旋转,这里由于是递归地执行,保证了平衡因子达到2的节点是最接近插入点的
if (key < tree->left->key)
tree = LL_Rotation(tree);
else
tree = LR_Rotation(tree);
}
}
else {
tree->right = insert(tree->right, key);
if (Height(tree->right) - Height(tree->left) >= ) {
if (key > tree->right->key)
tree = RR_Rotation(tree);
else
tree = RL_Rotation(tree);
}
}
tree->Height = Max(Height(tree->left), Height(tree->right)) + ;//当前节点的高度为其最大子树的高度+1
return tree;
} AVLTree LR_Rotation(AVLTree tree) {
tree->left = RR_Rotation(tree->left);
return LL_Rotation(tree);
} AVLTree RL_Rotation(AVLTree tree) {
tree->right = LL_Rotation(tree->right);
return RR_Rotation(tree);
} AVLTree RR_Rotation(AVLTree tree) {
AVLTree tree2 = tree->right;
tree->right = tree2->left;
tree2->left = tree;
tree->Height = Max(Height(tree->left), Height(tree->right)) + ;
tree2->Height = Max(Height(tree2->right), tree->Height) + ;
return tree2;
} AVLTree LL_Rotation(AVLTree tree) {
AVLTree tree2 = tree->left;
tree->left = tree2->right;
tree2->right = tree;
tree->Height = Max(Height(tree->left), Height(tree->right)) + ;
tree2->Height = Max(Height(tree->left), tree->Height) + ;
return tree2;
} int Height(AVLTree tree) {
if (tree == NULL)
return ;
return tree->Height;
} ElementType Max(ElementType a, ElementType b) {
return a > b ? a : b;
} void levelTraversal(AVLTree tree)
{
bool flag2 = false;
AVLTree t = NULL;
queue <AVLTree> Q;
Q.push(tree);
while (!Q.empty()) {
t = Q.front();
Q.pop();
cout << t->key;
if (flag2 && Height(t) != ) //高度为1的节点就是叶子节点
flag = false;
if (Height(t->left) < Height(t->right)) //AVL树保证了每个节点的左右子树高度差小于等于1,只要左子树高度小于右子树,这课AVL树就不是完全二叉树
flag = false;
if (t->left != NULL && t->right == NULL) //遇到一个节点,有左孩子但没有右孩子,标记一下,它之后的存入队列中的节点都为叶子节点时这棵AVL树才是完全二叉树
flag2 = true;
if (t->left != NULL)
Q.push(t->left);
if (t->right != NULL)
Q.push(t->right);
if (!Q.empty())
printf(" ");
}
printf("\n");
}
PAT甲级——1123 Is It a Complete AVL Tree (完全AVL树的判断)的更多相关文章
- PAT甲级1123. Is It a Complete AVL Tree
PAT甲级1123. Is It a Complete AVL Tree 题意: 在AVL树中,任何节点的两个子树的高度最多有一个;如果在任何时候它们不同于一个,则重新平衡来恢复此属性.图1-4说明了 ...
- PAT甲级1123 Is It a Complete AVL Tree【AVL树】
题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805351302414336 题意: 给定n个树,依次插入一棵AVL ...
- PAT甲级——A1123 Is It a Complete AVL Tree【30】
An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child sub ...
- PAT Advanced 1123 Is It a Complete AVL Tree (30) [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甲级1123
1123 Is It a Complete AVL Tree(30 分) An AVL tree is a self-balancing binary search tree. In an AVL t ...
- PAT甲级题解-1123. Is It a Complete AVL Tree (30)-AVL树+满二叉树
博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6806292.html特别不喜欢那些随便转载别人的原创文章又不给 ...
- PAT甲级题解-1066. Root of AVL Tree (25)-AVL树模板题
博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6803291.html特别不喜欢那些随便转载别人的原创文章又不给 ...
- 【PAT 甲级】1151 LCA in a Binary Tree (30 分)
题目描述 The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has bo ...
- PAT 甲级 1043 Is It a Binary Search Tree
https://pintia.cn/problem-sets/994805342720868352/problems/994805440976633856 A Binary Search Tree ( ...
随机推荐
- LLVM的总结
LLVM 写在前面的话:无意中看到的LLVM的作者Chris Lattner相关的介绍和故事,觉得很有意思就贴上来,如果不感兴趣,可以直接跳入下一章. 关于LLVM 如果你对LLVM的由来陌生,那么我 ...
- appium(6)-parts of appium api
parts of appium api Lock Lock the screen.//锁屏. // java driver.lockScreen(3); // objective c [driver ...
- CentOS(Linux) - 安装软件笔记(一) - VPSMate(服务器可视化管理工具)安装
更多详情点击参考官网说明 vpsmate官方推荐centos版本为6.2 64位 使用 SSH 连接工具,如 PuTTY.XShell.SecureCRT 等,连接到您的 Linux 服务器. 执行以 ...
- html5--4-4 audio元素/格式的转换
html5--4-4 audio元素/格式的转换 学习要点 掌握audio元素的用法 视频/音频文件的格式转换 当前,audio 元素支持三种音频格式: Ogg 免费, 支持的浏览器:Firefox. ...
- Ueditor文本编辑工具栏自定义
1.找到 ueditor.config.js 文件 2.在 toolbars 数组内进行删减,添加 , toolbars: [[ 'fullscreen', 'source', '|', 'undo' ...
- 关于 josnp ,解决跨域问题
JSONP是服务器与客户端跨源通信的常用方法.最大特点就是简单适用,老式浏览器全部支持,服务器不用做任何改造[使用jsonp的时候jsonp: "callback",callbac ...
- Opencv— — Twirl Filter
// define head function #ifndef PS_ALGORITHM_H_INCLUDED #define PS_ALGORITHM_H_INCLUDED #include < ...
- 「LuoguP3376」 【模板】网络最大流
题目描述 如题,给出一个网络图,以及其源点和汇点,求出其网络最大流. 输入输出格式 输入格式: 第一行包含四个正整数N.M.S.T,分别表示点的个数.有向边的个数.源点序号.汇点序号. 接下来M行每行 ...
- codevs 1497取余运算
1497 取余运算 时间限制: 1 s 空间限制: 128000 KB 题目等级 : 钻石 Diamon 题目描述 Description 输入b,p,k的值,编程计算bp mod k的值. ...
- MySQL源码学习——DBUG调试
一.前言 在规模稍微大点的项目中,为了方便快速找到bug的所在,我们往往需要在代码中加入一些调试用的代码,比如加入一些printf,打印出一些重点的信息:加入assert,进行断言判断.这些比较随意的 ...



