• 1066 Root of AVL Tree (模拟AVL建树)


    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 (≤) 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:

    70 

    Sample Input 2:

    7
    88 70 61 96 120 90 65

    Sample Output 2:

    88

     

    标准的模板题,为方便记忆,对7个函数的返回类型归纳如下:

    1个node* (newNode)

    2个int(getHeight, getBalance)

    4个void (updateHeight,L,R,insert )

    #include<cstdio>
    #include<algorithm>
    using namespace std;
    
    struct node{
        int data,height;
        node *left,*right;
    }*root;
    
    node* newNode(int x){
        node *ret=new node;
        ret->data=x;
        ret->height=1;
        ret->left=ret->right=NULL;
        return ret;
    }
    
    int getHeight(node *root){
        if(root==NULL) return 0;
        return root->height;
    }
    
    int getBalance(node *root){
        return getHeight(root->left)-getHeight(root->right);
    }
    
    void updateHeight(node *root){
        int hleft=getHeight(root->left);
        int hright=getHeight(root->right);
        root->height=max(hleft,hright)+1;
    }
    
    void L(node* &root){
        node *temp=root->right;
        root->right=temp->left;
        temp->left=root;
        updateHeight(root);
        updateHeight(temp);
        root=temp;
    }
    
    void R(node* &root){
        node *temp=root->left;
        root->left=temp->right;
        temp->right=root;
        updateHeight(root);
        updateHeight(temp);
        root=temp;
    }
    
    void insert(node* &root,int x){
        if(root==NULL){
            root=newNode(x);
            return;//空return更简洁,不用else 
        }
        if(x<root->data){
            insert(root->left,x);
            updateHeight(root);//* 
            if(getBalance(root)==2){
                if(getBalance(root->left)==1){
                    R(root);
                }else if(getBalance(root->left)==-1){
                    L(root->left);//* 
                    R(root);//* 
                }
            }
        }else{
            insert(root->right,x);
            updateHeight(root);
            if(getBalance(root)==-2){
                if(getBalance(root->right)==-1){
                    L(root);
                }else if(getBalance(root->right)==1){
                    R(root->right);
                    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",root->data);
        return 0;
    }
  • 相关阅读:
    Spring security中的BCryptPasswordEncoder方法对密码进行加密与密码匹配
    Eclipse导入SpringBoot项目pom.xml第一行报错Unknown error
    分库分表理论概述
    什么是乐观锁,什么是悲观锁
    oracle中的索引查看
    手动实现tail
    KNN理论
    矩阵以及向量
    numpy常用的几个小函数
    线性回归
  • 原文地址:https://www.cnblogs.com/exciting/p/10428935.html
Copyright © 2020-2023  润新知