一、题目描述
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:
70
Sample Input 2:
7
88 70 61 96 120 90 65
Sample Output 2:
88
二、解题思路
这道题是一道比较完整的AVL树的题目,基本涉及到了AVL树的所有操作,很适合学完AVL树之后用来巩固。具体的知识可以参考《算法笔记》中关于AVL树的讲解。
三、AC代码
#include<iostream>
#include<algorithm>
using namespace std;
struct Node //建立结点结构体
{
int key;
Node* lchild, *rchild;
};
Node* rotateLeft(Node* root) //左旋操作
{
Node *t = root->rchild;
root->rchild = t->lchild;
t->lchild = root;
return t;
}
Node* rotateRight(Node* root) //右旋操作
{
Node *t = root->lchild;
root->lchild = t->rchild;
t->rchild = root;
return t;
}
Node* rotateLeftRight(Node* root) //先左旋再右旋
{
root->lchild = rotateLeft(root->lchild);
return rotateRight(root);
}
Node* rotateRightLeft(Node* root) //先右旋再左旋
{
root->rchild = rotateRight(root->rchild);
return rotateLeft(root);
}
int getHeight(Node* root) //获取高度
{
if(root == NULL) return 0;
return max(getHeight(root->lchild), getHeight(root->rchild)) + 1;
}
Node* insert(Node *root, int val) //AVL树的插入
{
if(root == NULL)
{
root = new Node();
root->lchild = NULL;
root->rchild = NULL;
root->key = val;
return root;
}
else if(val < root->key)
{
root->lchild = insert(root->lchild, val);
if(getHeight(root->lchild) - getHeight(root->rchild) == 2)
root = val < root->lchild->key ? rotateRight(root) : rotateLeftRight(root);
}
else
{
root->rchild = insert(root->rchild, val);
if(getHeight(root->rchild) - getHeight(root->lchild) == 2)
root = val > root->rchild->key ? rotateLeft(root) : rotateRightLeft(root);
}
return root;
}
int main()
{
int N, val;
scanf("%d", &N);
Node *root = NULL;
for(int i=0; i<N; i++)
{
scanf("%d", &val);
root = insert(root, val);
}
printf("%d", root->key);
return 0;
}