可能的二叉搜索树和具有 n 个键的二叉树的总数

具有 n 个不同键的可能二叉搜索树的总数 (countBST(n)) = Catalan number Cn = (2n)！ /((n + 1)！* n！)
对于 n = 0, 1, 2, 3, … 加泰罗尼亚数字的值为 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …。二叉搜索树的数量也是如此。
具有 n 个不同键的可能二叉树的总数 (countBT(n)) = countBST(n) * n！

下面是用于查找具有 n 个数字的 BST 和二叉树的计数的代码。查找第 n 个加泰罗尼亚数字的代码取自此处。

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

考虑以每个元素为根的所有可能的二叉搜索树。如果有n个节点，那么对于每个选择的根节点，有n-1个非根节点，这些非根节点必须被划分为小于选定根的节点和大于选定根的节点.

假设节点 i 被选为根节点。然后有 i-1 个小于 i 的节点和 n-i 个大于 i 的节点。对于这两组节点中的每一个，都有一定数量的可能子树。

设 t(n) 是具有 n 个节点的 BST 的总数。以 i 为根的 BST 总数为 t(i – 1) t(n – i)。这两项相乘是因为左右子树中的排列是独立的。也就是说，对于左树中的每个排列和右树中的每个排列，您都会得到一个以 i 为根的 BST。

对 i 求和得到具有 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 BST，有一个 BST 有一个节点。

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