简单数据结构———AVL树

Posted 2020-07-23 当蜗牛有了理想

C - 万恶的二叉树

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2193

Description

An AVL tree is a kind of balanced binary search tree. Named after their inventors, Adelson-Velskii and Landis, they were the first dynamically balanced trees to be proposed. Like red-black trees, they are not perfectly balanced, but pairs of sub-trees differ in height by at most 1, maintaining an O(logn) search time. Addition and deletion operations also take O(logn) time.
Definition of an AVL tree
An AVL tree is a binary search tree which has the following properties:
1. The sub-trees of every node differ in height by at most one.
2. Every sub-tree is an AVL tree.

Balance requirement for an AVL tree: the left and right sub-trees differ by at most 1 in height.An AVL tree of n nodes can have different height.
For example, n = 7:

So the maximal height of the AVL Tree with 7 nodes is 3.
Given n,the number of vertices, you are to calculate the maximal hight of the AVL tree with n nodes.

 

Input

Input file contains multiple test cases. Each line of the input is an integer n(0<n<=10^9).
A line with a zero ends the input.

 

Output

An integer each line representing the maximal height of the AVL tree with n nodes.

 

Sample Input

1 2 0

 

Sample Output

0 1

题目大意：AVL树（又称高度平衡树）简介戳这里：http://baike.baidu.com/link?url=OUs8h211wSvjifeFg5cYzyUYmE8_WegNVk9MISMatdeLNDjMl1ypSn1CPUEnKIp1RQcFLFiuFfj4JwVOeOS0Sa

思路分析：本题只用到了AVL树的一个简单结论

高度为h的AVL树，节点数N最多为2^h-1,最少为N(h)=N(h-1)+N(h-2)+1

初始化N(0)=1,N(1)=2,以此类推

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
using namespace std;
const int maxn=50;;
char m[maxn][maxn];
__int64 a[maxn];
int main()
{
    a[0]=1;
    a[1]=2;
    __int64 n;
    for(int i=2;i<maxn;i++) a[i]=a[i-1]+a[i-2]+1;
    while(scanf("%I64d",&n)&&n)
    {
        int k=0;
        while(a[k]<=n) k++;
        cout<<--k<<endl;
    }
    return 0;
}