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

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!

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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.

 // See https://www.geeksforgeeks.org/program-nth-catalan-number/amp/  // 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 "<

 // See https://www.geeksforgeeks.org/program-nth-catalan-number/amp/  // 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 

 # 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) 

 // See https://www.geeksforgeeks.org/program-nth-catalan-number/amp/  // 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 

  $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  ?> 

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 contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.