# Total number of possible Binary Search Trees and Binary Trees with n keys

• Difficulty Level : Medium
• Last Updated : 16 Jun, 2022

Total number of possible Binary Search Trees with n different keys (countBST(n)) = Catalan number Cn = (2n)! / ((n + 1)! * n!)

For n = 0, 1, 2, 3, … values of Catalan numbers are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …. So are numbers of Binary Search Trees.

Total number of possible Binary Trees with n different keys (countBT(n)) = countBST(n) * n!

Below is code for finding count of BSTs and Binary Trees with n numbers. The code to find n’th Catalan number is taken from here.

## 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 numberunsigned 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) timeunsigned 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 catalanunsigned 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 nodesunsigned 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 functionsint 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 numberstatic 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) timestatic 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 catalanstatic 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 nodesstatic int countBT(int n){    // find count of BST    // with n numbers    int count = catalan(n);     // return count * n!    return count * factorial(n);} // Driver Codepublic 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 numberdef 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) timedef 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 catalandef 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 nodesdef countBT(n):     # find count of BST with n numbers    count = catalan(n)     # return count * n!    return count * factorial(n) # Driver Codeif __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 numberstatic 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) timestatic 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 catalanstatic 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 nodesstatic int countBT(int n){    // find count of BST    // with n numbers    int count = catalan(n);     // return count * n!    return count * factorial(n);} // Driver Codestatic 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 $res;} // A Binomial coefficient// based function to find// nth catalan number in// O(n) timefunction catalan($n){ // Calculate value of 2nCn $c = binomialCoeff(2 * $n, $n);     // return 2nCn/(n+1)    return (int)$c / ($n + 1);} // A function to count// number of BST with// n nodes using catalanfunction countBST($n){ // find nth catalan number $count = catalan($n);  // return nth // catalan number return $count;} // A function to count// number of binary// trees with n nodesfunction countBT($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 nodesecho "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

 

Output:

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

Proof of Enumeration

Consider all possible binary search trees with each element at the root. If there are n nodes, then for each choice of root node, there are n – 1 non-root nodes and these non-root nodes must be partitioned into those that are less than a chosen root and those that are greater than the chosen root.

Let’s say node i is chosen to be the root. Then there are i – 1 nodes smaller than i and n – i nodes bigger than i. For each of these two sets of nodes, there is a certain number of possible subtrees.

Let t(n) be the total number of BSTs with n nodes. The total number of BSTs with i at the root is t(i – 1) t(n – i). The two terms are multiplied together because the arrangements in the left and right subtrees are independent. That is, for each arrangement in the left tree and for each arrangement in the right tree, you get one BST with i at the root.

Summing over i gives the total number of binary search trees with n nodes. The base case is t(0) = 1 and t(1) = 1, i.e. there is one empty BST and there is one BST with one node.   Also, the relationship countBT(n) = countBST(n) * n! holds. As for every possible BST, there can have n! binary trees where n is the number of nodes in BST.

This article is contributed by Shubham Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.