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++
// for reference of below code.#include <bits/stdc++.h>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 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 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 "<<n<<" nodes is "<<count1<<endl; cout<<"Count of binary trees with "<<n<<" nodes is "<<count2; return 0;} |
Java
// 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 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 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 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#
// 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 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 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
<?php// for reference of below code.// A function to find factorial// of a given numberfunction factorial($n){ $res = 1; // Calculate value of // [1*(2)*---*(n-k+1)] / // [k*(k-1)*---*1] for ($i = 1; $i <= $n; ++$i) { $res *= $i; } return $res;}function binomialCoeff($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 = 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 $count;}// A function to count // number of binary // trees with n nodes function countBT($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 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
<script>// 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); </script> |
Output:
Count of BST with 5 nodes is 42 Count of binary trees with 5 nodes is 5040
Time Complexity: O(n): The time complexity of the above code is O(n). It uses the Catalan number formula to calculate the number of possible binary search trees in O(n) time.
Auxiliary Space Complexity: O(1), The space complexity of the above code is O(1). No extra space is allocated.
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.
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