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
- cgic 中文文档
CGIC英文文档地址:https://boutell.com/cgic/ cgic是用c写cgi程序的一个很小的库,所以英文文档也很少,为了便于日后复习翻看,心血来潮,翻译了一遍. 1. 什么是cgi ...
- linux内核段属性机制【转】
本文转载自:https://github.com/TongxinV/oneBook/issues/9 linux内核段属性机制 以subsys_initcall和module_init为例 subsy ...
- Unable to create Debug Bridge:Unable to start adb server:error:cannot parse version
打开Android Studio时报如下错误提示: Unable to create Debug Bridge:Unable to start adb server:error:cannot pars ...
- css元素定位样式
曾经写网页,学css整体上不难,但就是元素定位,始终一知半解,直到今天,本着实践出真知的理念,经过认真测试,总结出了如下结论. css 定位: positionstatic : 默认静止定位,元素在正 ...
- ubuntu166.04之Caffe安装
写在前面:之前一直在搞keras,最近由于某些需求,需要学习caffe,在此记录caffe的安装记录.默认已经安装了cuda 如果是从其他的深度学习平台迁移到Caffe,那么按照这个教程来就可以了. ...
- LuoguP4861 按钮
传送门 这题一眼看上去要解\(k^x \equiv 1(mod\ m)\)的最小正整数解. 于是我打了一个扩展BSGS 这题这样做算的答案一直是0的.不过有另一个定理欧拉定理,\(k^{\varphi ...
- No java virtual machine ....
运行Eclipse提示No java virtual machine 版权声明:本文原创作者:一叶飘舟 作者博客地址:http://blog.csdn.net/jdsjlzx http://blo ...
- Java使用Jacob转换Word为HTML
从今天开始,我也要养成记录开发中遇到的问题和解决方法的好习惯! 最近开发一个Android项目,需要用到查看Word和Pdf文档的功能,由于Android没有直接显示Word和PDF文档的组件,只有一 ...
- Spring Boot配置多个DataSource
使用Spring Boot时,默认情况下,配置DataSource非常容易.Spring Boot会自动为我们配置好一个DataSource. 百牛信息技术bainiu.ltd整理发布于博客园 如果在 ...
- FFmpeg在Linux下搭建 ***
今天介绍下FFmpeg在Linux下安装编译过程,总体过程比较顺利,就是在ffmpeg等的时间稍长点,仅当记录. 关于FFmpeg FFmpeg是一个开源免费跨平台的视频和音频流方案,属于自由软件,采 ...