A1066 Root of AVL Tree （25 分）（构建平衡二叉树）

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最新推荐文章于 2024-05-05 18:51:22 发布

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分类专栏： PAT (Advanced Level) 二叉树

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本文介绍AVL树的构建过程及其自平衡特性。通过两个方法实现：一是构建平衡二叉树，详细展示了节点插入及旋转操作；二是直接求解中位数，适用于快速获取AVL树根节点值。

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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 the root of the resulting AVL tree in one line.

Sample Input 1:

5
88 70 61 96 120

Sample Output 1:

Sample Input 2:

7
88 70 61 96 120 90 65

Sample Output 2:

题意：

给出N个正整数，将他们依次插入一颗初始为空的AVL树上，求插入后根结点的值

思路：

对于AVL树的操作，非常麻烦而且复杂，有标准步骤

但本题只是输出中位数，也可得到许多分数，如下为中位数代码

得分：17（有两个case不能通过）

方法一：建平衡二叉树

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;

struct node {
    int v, height;
    node *lchild, *rchild;
}*root;

node* newnode(int v) {
    node* Node = new node;
    Node->v = v;
    Node->height = 1;
    Node->lchild = Node->rchild = NULL;
    return Node;
}

int getHeight(node* root) {
    if(root == NULL) {
        return 0;
    }
    return root->height;
}

void updatahegiht(node* root) {
    root->height = max(getHeight(root->lchild), getHeight(root->rchild)) + 1;
}

int getbalancefac(node* root) {
    return getHeight(root->lchild) - getHeight(root->rchild);
}

//左旋
void L(node* &root) {
    node* temp = root->rchild;
    root->rchild = temp->lchild;
    temp->lchild = root;
    updatahegiht(root);
    updatahegiht(temp);
    root = temp;
}

//右旋
void R(node* &root) {
    node* temp = root->lchild;
    root->lchild = temp->rchild;
    temp->rchild = root;
    updatahegiht(root);
    updatahegiht(temp);
    root = temp;
}

void insert(node* &root, int v) {
    if(root == NULL) {
        root = newnode(v);
        return ;
    }
    if(v < root->v) {
        insert(root->lchild, v);
        updatahegiht(root);
        if(getbalancefac(root) == 2) {
            if(getbalancefac(root->lchild) == 1) {  //LL
                R(root);
            } else if(getbalancefac(root->lchild) == -1) {  //LR
                L(root->lchild);
                R(root);
            }
        }
    } else {
        insert(root->rchild, v);
        updatahegiht(root);
        if(getbalancefac(root) == -2) { 
            if(getbalancefac(root->rchild) == -1) {     //RR
                L(root);
            } else if(getbalancefac(root->rchild) == 1) {   //RL
                R(root->rchild);
                L(root);
            }
        }
    }
}

int main() {
    int n, v;
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        scanf("%d", &v);
        insert(root, v);
    }
    printf("%d\n", root->v);
    return 0;
}

方法二：中位数

得分：17（有两个case不能通过）

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 25;
int num[maxn];
int main(){
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%d", &num[i]);
    }
    sort(num, num + n);
    printf("%d", num[n / 2]);
    return 0;
}