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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 【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 ...

  9. 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

  1. Linux就该这么学--命令集合1(常用系统工作命令)

    1.用echo命令查看SHELL变量的值(前面有$符号): echo $SHELL 2.查看本机主机名: echo $HOSTNAME 3.查看当前的系统时间: date 4.按照“年-月-日 时:分 ...

  2. RecyclerView 可以与CollapsingToolbarLayout一起使用

    Item  布局 <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:and ...

  3. zsh 的简单介绍

    什么是 zsh,要想解释好这个问题,那么得先说明什么是 shell.不负责任的解释说法就是 shell 就是一个壳.这个壳可不是蜗牛的壳,而是计算机的一个壳,当然也不是计算机的外壳啦,这个壳是相对于计 ...

  4. linux内核container_of宏定义分析

    看见一个哥们分析container_of很好,转来留给自己看 一.#define offsetof(TYPE, MEMBER) ((size_t) & ((TYPE *)0)->MEMB ...

  5. Android项目中JNI技术生成并调用.so动态库实现详解

    生成 jni方式有两种:一种是通过SWIG从C++代码生成过度的java代码:另一种是通过javah的方式从java代码自动生成过度的C++代码.两种方式下的步骤流程正好相反. 第一种方式:由于需要配 ...

  6. bzoj4485: [Jsoi2015]圈地

    思维僵化选手在线被虐 其实应该是不难的,题目明显分成两个集合,要求是不同集合的点不能联通 先假设全选了,然后二分图最小割,相邻两个点直接连墙的费用就可以了 #include<cstdio> ...

  7. HDU2586 How far away? —— 倍增LCA

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2586 How far away ? Time Limit: 2000/1000 MS (Java/Ot ...

  8. ansible 文件模块,很实用

    摘自: http://blog.csdn.net/kellyseeme/article/details/50545521

  9. mongodb给我们提供了fsync+lock机制把数据暴力的刷到硬盘上

    能不能把数据暴力的刷到硬盘上,当然是可以的,mongodb给我们提供了fsync+lock机制就能满足我们提的需求. fsync+lock首先会把缓冲区数据暴力刷入硬盘,然后给数据库一个写入锁,其他实 ...

  10. AndroidManifest中的Intent-filter标签

    经过测试,intent-filter标签中的: 1. <action android:name="android.intent.action.MAIN" /> 代表这是 ...