先决条件:带有n个键的可能的二进制搜索树总数
给定N个整数的数组arr [] 。任务是计算可以使用arr []中元素的每个节点作为根节点进行二进制搜索树的数量。
例子:
Input: arr[] = { 20, 10, 30 }
Output: 1 2 2
Input: arr[] = { 1, 2, 3, 4, 5 }
Output: 14 5 4 5 14
方法:
二进制搜索树(BST)的总数由加泰罗尼亚语数字给出:
Cn = (2n)!/(( n+1)!*n!)
where n = number of distinct keys.
- 计算少于当前节点的元素数量(例如c1 )。
- 计算大于当前节点的元素数量(例如c2 )。
- 然后,可以使用当前元素作为根节点来形成二叉搜索树(BST)的总数等于使用c1个元素形成的BST总数与使用c2个元素形成的BST总数的乘积。
Total Number of BST = Cc1*Cc2
下面是上述方法的实现:
C++
// C++ implementation of the above code
#include
using namespace std;
// A function to find factorial of a
// given number
int fact(int num)
{
int fact = 1;
while(num > 1)
{
fact *= num;
num -= 1;
}
return fact;
}
// Find nth catalan number
int catalan(int n)
{
return fact(2 * n)/(fact(n) * fact(n + 1)) ;
}
// Driver Code
int main()
{
// size of arr[]
int n = 5;
// Elements in arr[]
int arr[] = {1, 2, 3, 4, 5};
int i,k;
for(k = 0; k < n; k++)
{
int s = 0;
// Count the number of element
// less than current element
// in arr[k]
for(i = 0; i < n; i++)
{
if (arr[i] < arr[k])
s += 1 ;
}
// Here s = number of node in left
// BST and (n-s-1) = number of node
// in right BST
// Find number of BST using elements
// in left BST
int catalan_leftBST = catalan(s) ;
// Find number of BST using elements
// in right BST
int catalan_rightBST = catalan(n - s - 1) ;
// Find total number of BST
int totalBST = catalan_rightBST * catalan_leftBST ;
// Print total BST count
cout<< totalBST << " " ;
}
}
// This code is contributed by AnkitRai01
Java
// java implementation of the above code
public class GFG
{
// A function to find factorial of a
// given number
static int fact(int num)
{
int fact = 1;
while(num > 1)
{
fact *= num;
num -= 1;
}
return fact;
}
// Find nth catalan number
static int catalan(int n)
{
return fact(2 * n)/(fact(n) * fact(n + 1)) ;
}
// Driver Code
public static void main (String[] args)
{
// size of arr[]
int n = 5;
// Elements in arr[]
int arr[] = {1, 2, 3, 4, 5};
int i,k;
for(k = 0; k < n; k++)
{
int s = 0;
// Count the number of element
// less than current element
// in arr[k]
for(i = 0; i < n; i++)
{
if (arr[i] < arr[k])
s += 1 ;
}
// Here s = number of node in left
// BST and (n-s-1) = number of node
// in right BST
// Find number of BST using elements
// in left BST
int catalan_leftBST = catalan(s) ;
// Find number of BST using elements
// in right BST
int catalan_rightBST = catalan(n - s - 1) ;
// Find total number of BST
int totalBST = catalan_rightBST * catalan_leftBST ;
// Print total BST count
System.out.print(totalBST + " ") ;
}
}
}
// This code is contributed by AnkitRai01
Python3
# A function to find factorial of a
# given number
def fact(num):
fact = 1;
while(num>1):
fact = fact * num;
num = num-1;
return fact;
# Find nth catalan number
def catalan(n):
return fact(2 * n)//(fact(n)*fact(n + 1))
# Driver Code
if __name__ == '__main__':
# size of arr[]
n = 5
# Elements in arr[]
arr = [1, 2, 3, 4, 5]
for k in range(n):
s = 0
# Count the number of element
# less than current element
# in arr[k]
for i in range(n):
if arr[i] < arr[k]:
s+= 1
# Here s = number of node in left
# BST and (n-s-1) = number of node
# in right BST
# Find number of BST using elements
# in left BST
catalan_leftBST = catalan(s)
# Find number of BST using elements
# in right BST
catalan_rightBST = catalan(n-s-1)
# Find total number of BST
totalBST = catalan_rightBST * catalan_leftBST
# Print total BST count
print(totalBST, end =" ")
C#
// C# implementation of the above code
using System;
class GFG
{
// A function to find factorial of a
// given number
static int fact(int num)
{
int fact = 1;
while(num > 1)
{
fact *= num;
num -= 1;
}
return fact;
}
// Find nth catalan number
static int catalan(int n)
{
return fact(2 * n)/(fact(n) * fact(n + 1)) ;
}
// Driver Code
public static void Main(String[] args)
{
// size of []arr
int n = 5;
// Elements in []arr
int []arr = {1, 2, 3, 4, 5};
int i,k;
for(k = 0; k < n; k++)
{
int s = 0;
// Count the number of element
// less than current element
// in arr[k]
for(i = 0; i < n; i++)
{
if (arr[i] < arr[k])
s += 1 ;
}
// Here s = number of node in left
// BST and (n-s-1) = number of node
// in right BST
// Find number of BST using elements
// in left BST
int catalan_leftBST = catalan(s) ;
// Find number of BST using elements
// in right BST
int catalan_rightBST = catalan(n - s - 1) ;
// Find total number of BST
int totalBST = catalan_rightBST * catalan_leftBST ;
// Print total BST count
Console.Write(totalBST + " ") ;
}
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
14 5 4 5 14
时间复杂度: O(N 2 )