Given an integer n, the task is to find the total number of unique Binary Search trees And unique Binary trees that can be made using values from 1 to n.
Examples:
Input: n = 3
Output: BST = 5
BT = 30
Explanation: For n = 3, Total number of binary search tree will be 5 and total Binary tree will b 30.Input: n = 5
Output: BST = 42
BT = 5040
Explanation: For n = 5, Total number of binary search tree will be 42 and total Binary tree will b 5040.
Approach:
1. Unique Binary Search Trees (BSTs):
First let's see how we find Total number of binary search tree with n nodes. A binary search tree (BST) maintains the property that elements are arranged based on their relative order. Let’s define C(n) as the number of unique BSTs that can be constructed with
nnodes. This is given by the following recursive formula:
- C(n) = Σ(i = 1 to n) C(i-1) * C(n-i).
This formula corresponds to the recurrence relation for the nth Catalan number. Please refer to Number of Unique BST with N Keys for better understanding and proof. We just need to find nth catalan number. First few catalan numbers are 1 1 2 5 14 42 132 429 1430 4862, … (considered from 0th number).
- Formula of catalan number is (1 / n+1) * ( 2*nCn). Please refer to Applications of Catalan Numbers.
2. Unique Binary Trees (General Binary Trees):
For general binary trees, the nodes do not have to follow the Binary Search Tree property. The total number of unique binary trees is calculated as: Total Binary Trees = countBST(n) * n!
// C++ code of finding Number of Unique
// BST and BT with N Keys
#include <iostream>
using namespace std;
// Function to calculate the binomial
// coefficient C(n, k)
int binomialCoeff(int n, int k) {
// C(n, k) is the same as C(n, n-k)
if (k > n - k) {
k = n - k;
}
int res = 1;
// Calculate the value of n! / (k! * (n-k)!)
for (int i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
// Function to find the nth Catalan number
int numBST(int n) {
// Calculate C(2n, n)
int c = binomialCoeff(2 * n, n);
// Return the nth Catalan number
return c / (n + 1);
}
// Function to find total binary tree
int numBT(int n) {
int fact = 1;
// Calculating factorial
for(int i = 1; i <= n; i++) {
fact = fact * i;
}
// Total binary tree = Tatal binary search tree * n!
return numBST(n) * fact;
}
int main() {
int n = 5;
cout << numBST(n) << endl;
cout << numBT(n) << endl;
return 0;
}
// Java code of finding Number of Unique
// BST and BT with N Keys
class GfG {
// Function to calculate the binomial coefficient C(n, k)
static int binomialCoeff(int n, int k) {
// C(n, k) is the same as C(n, n-k)
if (k > n - k) {
k = n - k;
}
int res = 1;
// Calculate the value of n! / (k! * (n-k)!)
for (int i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
// Function to find the nth Catalan number
static int numBST(int n) {
// Calculate C(2n, n)
int c = binomialCoeff(2 * n, n);
// Return the nth Catalan number
return c / (n + 1);
}
// Function to find total binary tree
static int numBT(int n) {
int fact = 1;
// Calculating factorial
for (int i = 1; i <= n; i++) {
fact = fact * i;
}
// Total binary tree = Total binary search tree * n!
return numBST(n) * fact;
}
public static void main(String[] args) {
int n = 5;
System.out.println(numBST(n));
System.out.println(numBT(n));
}
}
# Python code of finding Number of Unique
# BST and BT with N Keys
# Function to calculate the binomial coefficient C(n, k)
def binomialCoeff(n, k):
# C(n, k) is the same as C(n, n-k)
if k > n - k:
k = n - k
res = 1
# Calculate the value of n! / (k! * (n-k)!)
for i in range(k):
res *= (n - i)
res //= (i + 1)
return res
# Function to find the nth Catalan number
def numBST(n):
# Calculate C(2n, n)
c = binomialCoeff(2 * n, n)
# Return the nth Catalan number
return c // (n + 1)
# Function to find total binary tree
def numBT(n):
fact = 1
# Calculating factorial
for i in range(1, n + 1):
fact *= i
# Total binary tree = Total binary search tree * n!
return numBST(n) * fact
n = 5
print(numBST(n))
print(numBT(n))
// C# code of finding Number of Unique
// BST and BT with N Keys
using System;
class GfG {
// Function to calculate the binomial coefficient C(n, k)
static int binomialCoeff(int n, int k) {
// C(n, k) is the same as C(n, n-k)
if (k > n - k) {
k = n - k;
}
int res = 1;
// Calculate the value of n! / (k! * (n-k)!)
for (int i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
// Function to find the nth Catalan number
static int numBST(int n) {
// Calculate C(2n, n)
int c = binomialCoeff(2 * n, n);
// Return the nth Catalan number
return c / (n + 1);
}
// Function to find total binary tree
static int numBT(int n) {
int fact = 1;
// Calculating factorial
for (int i = 1; i <= n; i++) {
fact = fact * i;
}
// Total binary tree = Total binary
// search tree * n!
return numBST(n) * fact;
}
static void Main() {
int n = 5;
Console.WriteLine(numBST(n));
Console.WriteLine(numBT(n));
}
}
// JavaScript code of finding Number of Unique
// BST and BT with N Keys
// Function to calculate the binomial coefficient C(n, k)
function binomialCoeff(n, k) {
// C(n, k) is the same as C(n, n-k)
if (k > n - k) {
k = n - k;
}
let res = 1;
// Calculate the value of n! / (k! * (n-k)!)
for (let i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
// Function to find the nth Catalan number
function numBST(n) {
// Calculate C(2n, n)
let c = binomialCoeff(2 * n, n);
// Return the nth Catalan number
return c / (n + 1);
}
// Function to find total binary tree
function numBT(n) {
let fact = 1;
// Calculating factorial
for (let i = 1; i <= n; i++) {
fact *= i;
}
// Total binary tree = Total binary search tree * n!
return numBST(n) * fact;
}
let n = 5;
console.log(numBST(n));
console.log(numBT(n));
Output
42 5040
Time Complexity: O(n), where n is total number of node
Auxiliary Space: O(1)
