可能的二叉搜索树和具有n个键的二叉树的总数

具有n个不同关键字(countBST(n))的可能二进制搜索树总数=加泰罗尼亚语数量Cn =(2n)！ /((n + 1)！* n！)
对于n = 0、1、2、3，…，加泰罗尼亚语值的值为1、1、2、5、14、42、132、429、1430、4862等。二叉搜索树的数量也是如此。
具有n个不同密钥的可能二叉树总数(countBT(n))= countBST(n)* n！

下面的代码用于查找具有n个数字的BST和二叉树的计数。查找第n个加泰罗尼亚语代码的代码是从此处获取的。

C++

// See https://www.geeksforgeeks.org/program-nth-catalan-number/
// for reference of below code.
 
#include 
using namespace std;
 
// A function to find factorial of a given number
unsigned long int factorial(unsigned int n)
{
    unsigned long int res = 1;
 
    // Calculate value of [1*(2)*---*(n-k+1)] / [k*(k-1)*---*1]
    for (int i = 1; i <= n; ++i)
    {
        res *= i;
    }
 
    return res;
}
 
unsigned long int binomialCoeff(unsigned int n, unsigned int k)
{
    unsigned long int res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of [n*(n-1)*---*(n-k+1)] / [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
 
// A Binomial coefficient based function to find nth catalan
// number in O(n) time
unsigned long int catalan(unsigned int n)
{
    // Calculate value of 2nCn
    unsigned long int c = binomialCoeff(2*n, n);
 
    // return 2nCn/(n+1)
    return c/(n+1);
}
 
// A function to count number of BST with n nodes
// using catalan
unsigned long int countBST(unsigned int n)
{
    // find nth catalan number
    unsigned long int count = catalan(n);
 
    // return nth catalan number
    return count;
}
 
// A function to count number of binary trees with n nodes
unsigned long int countBT(unsigned int n)
{
    // find count of BST with n numbers
    unsigned long int count = catalan(n);
 
    // return count * n!
    return count * factorial(n);
}
 
// Driver Program to test above functions
int main()
{
 
    int count1,count2, n = 5;
 
    // find count of BST and binary trees with n nodes
        count1 = countBST(n);
        count2 = countBT(n);
     
    // print count of BST and binary trees with n nodes
    cout<<"Count of BST with "<

---

Java

// See https://www.geeksforgeeks.org/program-nth-catalan-number/
// for reference of below code.
import java.io.*;
 
class GFG
{
     
// A function to find
// factorial of a given number
static int factorial(int n)
{
    int res = 1;
 
    // Calculate value of
    // [1*(2)*---*(n-k+1)] /
    // [k*(k-1)*---*1]
    for (int i = 1; i <= n; ++i)
    {
        res *= i;
    }
 
    return res;
}
 
static int binomialCoeff(int n,
                         int k)
{
    int res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] /
    // [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
 
// A Binomial coefficient
// based function to find
// nth catalan number in
// O(n) time
static int catalan( int n)
{
     
    // Calculate value of 2nCn
    int c = binomialCoeff(2 * n, n);
 
    // return 2nCn/(n+1)
    return c / (n + 1);
}
 
// A function to count number of
// BST with n nodes using catalan
static int countBST( int n)
{
    // find nth catalan number
    int count = catalan(n);
 
    // return nth catalan number
    return count;
}
 
// A function to count number
// of binary trees with n nodes
static int countBT(int n)
{
    // find count of BST
    // with n numbers
    int count = catalan(n);
 
    // return count * n!
    return count * factorial(n);
}
 
// Driver Code
public static void main (String[] args)
{
    int count1, count2, n = 5;
 
    // find count of BST and
    // binary trees with n nodes
    count1 = countBST(n);
    count2 = countBT(n);
 
    // print count of BST and
    // binary trees with n nodes
    System.out.println("Count of BST with "+
                            n +" nodes is "+
                                    count1);
    System.out.println("Count of binary " +
                             "trees with "+
                         n + " nodes is " +
                                   count2);
}
}
 
// This code is contributed by ajit

Python3

# See https:#www.geeksforgeeks.org/program-nth-catalan-number/
# for reference of below code.
 
# A function to find factorial of a given number
def factorial(n) :
    res = 1
     
    # Calculate value of [1*(2)*---*
    #(n-k+1)] / [k*(k-1)*---*1]
    for i in range(1, n + 1):
        res *= i
    return res
 
def binomialCoeff(n, k):
 
    res = 1
 
    # Since C(n, k) = C(n, n-k)
    if (k > n - k):
        k = n - k
 
    # Calculate value of [n*(n-1)*---*(n-k+1)] /
    # [k*(k-1)*---*1]
    for i in range(k):
     
        res *= (n - i)
        res //= (i + 1)
     
    return res
 
# A Binomial coefficient based function to
# find nth catalan number in O(n) time
def catalan(n):
 
    # Calculate value of 2nCn
    c = binomialCoeff(2 * n, n)
 
    # return 2nCn/(n+1)
    return c // (n + 1)
 
# A function to count number of BST
# with n nodes using catalan
def countBST(n):
 
    # find nth catalan number
    count = catalan(n)
 
    # return nth catalan number
    return count
 
# A function to count number of binary
# trees with n nodes
def countBT(n):
 
    # find count of BST with n numbers
    count = catalan(n)
 
    # return count * n!
    return count * factorial(n)
 
# Driver Code
if __name__ == '__main__':
 
    n = 5
 
    # find count of BST and binary
    # trees with n nodes
    count1 = countBST(n)
    count2 = countBT(n)
 
    # print count of BST and binary trees with n nodes
    print("Count of BST with", n, "nodes is", count1)
    print("Count of binary trees with", n,
                       "nodes is", count2)
 
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#

// See https://www.geeksforgeeks.org/program-nth-catalan-number/
// for reference of below code.
using System;
 
class GFG
{
     
// A function to find
// factorial of a given number
static int factorial(int n)
{
    int res = 1;
 
    // Calculate value of
    // [1*(2)*---*(n-k+1)] /
    // [k*(k-1)*---*1]
    for (int i = 1; i <= n; ++i)
    {
        res *= i;
    }
 
    return res;
}
 
static int binomialCoeff(int n,
                         int k)
{
    int res = 1;
 
    // Since C(n, k) = C(n, n-k)
    if (k > n - k)
        k = n - k;
 
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] /
    // [k*(k-1)*---*1]
    for (int i = 0; i < k; ++i)
    {
        res *= (n - i);
        res /= (i + 1);
    }
 
    return res;
}
 
// A Binomial coefficient
// based function to find
// nth catalan number in
// O(n) time
static int catalan(int n)
{
     
    // Calculate value
    // of 2nCn
    int c = binomialCoeff(2 * n, n);
 
    // return 2nCn/(n+1)
    return c / (n + 1);
}
 
// A function to count
// number of BST with
// n nodes using catalan
static int countBST(int n)
{
    // find nth catalan number
    int count = catalan(n);
 
    // return nth catalan number
    return count;
}
 
// A function to count number
// of binary trees with n nodes
static int countBT(int n)
{
    // find count of BST
    // with n numbers
    int count = catalan(n);
 
    // return count * n!
    return count * factorial(n);
}
 
// Driver Code
static public void Main ()
{
    int count1, count2, n = 5;
     
    // find count of BST 
    // and binary trees
    // with n nodes
    count1 = countBST(n);
    count2 = countBT(n);
     
    // print count of BST and
    // binary trees with n nodes
    Console.WriteLine("Count of BST with "+
                           n +" nodes is "+
                                   count1);
    Console.WriteLine("Count of binary " +
                            "trees with "+
                        n + " nodes is " +
                                   count2);
    }
}
 
// This code is contributed
// by akt_mit

PHP

 $n - $k)
        $k = $n - $k;
 
    // Calculate value of
    // [n*(n-1)*---*(n-k+1)] /
    // [k*(k-1)*---*1]
    for ($i = 0; $i < $k; ++$i)
    {
        $res *= ($n - $i);
        $res = (int)$res / ($i + 1);
    }
 
    return $res;
}
 
// A Binomial coefficient
// based function to find
// nth catalan number in
// O(n) time
function catalan($n)
{
    // Calculate value of 2nCn
    $c = binomialCoeff(2 * $n, $n);
 
    // return 2nCn/(n+1)
    return (int)$c / ($n + 1);
}
 
// A function to count
// number of BST with
// n nodes using catalan
function countBST($n)
{
    // find nth catalan number
    $count = catalan($n);
 
    // return nth
    // catalan number
    return $count;
}
 
// A function to count
// number of binary
// trees with n nodes
function countBT($n)
{
    // find count of
    // BST with n numbers
    $count = catalan($n);
 
    // return count * n!
    return $count *
           factorial($n);
}
 
// Driver Code
$count1;
$count2;
$n = 5;
 
// find count of BST and
// binary trees with n nodes
$count1 = countBST($n);
$count2 = countBT($n);
 
// print count of BST and
// binary trees with n nodes
echo "Count of BST with " , $n ,
     " nodes is ", $count1,"\n";
      
echo "Count of binary trees with " ,
           $n ," nodes is ",$count2;
 
// This code is contributed by ajit
?>

Javascript

 
// See https://www.geeksforgeeks.org/program-nth-catalan-number/
// for reference of below code.
 
    // A function to find
    // factorial of a given number
    function factorial(n)
    {
        let res = 1;
 
        // Calculate value of
        // [1*(2)*---*(n-k+1)] /
        // [k*(k-1)*---*1]
        for (let i = 1; i <= n; ++i)
        {
            res *= i;
        }
 
        return res;
    }
 
    function binomialCoeff(n, k)
    {
        let res = 1;
 
        // Since C(n, k) = C(n, n-k)
        if (k > n - k)
            k = n - k;
 
        // Calculate value of
        // [n*(n-1)*---*(n-k+1)] /
        // [k*(k-1)*---*1]
        for (let i = 0; i < k; ++i)
        {
            res *= (n - i);
            res /= (i + 1);
        }
 
        return res;
    }
 
    // A Binomial coefficient
    // based function to find
    // nth catalan number in
    // O(n) time
    function catalan(n)
    {
 
        // Calculate value
        // of 2nCn
        let c = binomialCoeff(2 * n, n);
 
        // return 2nCn/(n+1)
        return c / (n + 1);
    }
 
    // A function to count
    // number of BST with
    // n nodes using catalan
    function countBST(n)
    {
        // find nth catalan number
        let count = catalan(n);
 
        // return nth catalan number
        return count;
    }
 
    // A function to count number
    // of binary trees with n nodes
    function countBT(n)
    {
        // find count of BST
        // with n numbers
        let count = catalan(n);
 
        // return count * n!
        return count * factorial(n);
    }
       
    let count1, count2, n = 5;
       
    // find count of BST 
    // and binary trees
    // with n nodes
    count1 = countBST(n);
    count2 = countBT(n);
       
    // print count of BST and
    // binary trees with n nodes
    document.write("Count of BST with "+
                           n +" nodes is "+
                                   count1 + "</br>");
    document.write("Count of binary " +
                            "trees with "+
                        n + " nodes is " +
                                   count2);
                                    

输出：

Count of BST with 5 nodes is 42
Count of binary trees with 5 nodes is 5040

枚举证明

考虑所有可能的二进制搜索树，其中每个元素都位于根目录。如果有n个节点，那么对于每个根节点选择，都会有n – 1个非根节点，并且必须将这些非根节点划分为小于所选根节点的那些和大于所选根节点的那些。

假设节点i被选择为根。然后有i个节点-小于i个节点和n个-i个节点大于i个节点。对于这两组节点中的每一个，都有一定数量的可能子树。

令t(n)为具有n个节点的BST的总数。以i为根的BST总数为t(i – 1)t(n – i)。由于左和右子树中的排列是独立的，所以这两个项相乘在一起。也就是说，对于左树中的每个排列和右树中的每个排列，您都将获得一个BST，其中i为根。

对i求和得出具有n个节点的二叉搜索树的总数。

$t(n) = \sum_{i=1}^{n} t(i-1) t(n-i)$

基本情况是t(0)= 1和t(1)= 1，即有一个空的BST，并且有一个带有一个节点的BST。