The process in which a function calls itself directly or indirectly is called Recursion and the corresponding function is called a Recursive function.
Using Recursion, certain problems can be solved quite easily. Examples of such problems are Towers of Hanoi (TOH), Inorder/Preorder/Postorder Tree Traversals, DFS, etc.
Types of Recursions:
Recursion can be further classified into two kinds, depending on when they terminate:
- Finite Recursion
- Infinite Recursion
Finite Recursion:
Finite Recursion occurs when the recursion terminates after a finite number of recursive calls. A recursion terminates only when a base condition is met.
Example:
Below is an implementation to demonstrate Finite Recursion.
// C++ program to demsonstrate Finite Recursion #include <bits/stdc++.h> using namespace std;
// Recursive function void Geek( int N)
{ // Base condition
// When this condition is met,
// the recursion terminates
if (N == 0)
return ;
// Print the current value of N
cout << N << " " ;
// Call itself recursively
Geek(N - 1);
} // Driver code int main()
{ // Initial value of N
int N = 5;
// Call the recursive function
Geek(N);
return 0;
} |
// Java program for the above approach class GFG{
// Recursive function static void Geek( int N)
{ // Base condition
// When this condition is met,
// the recursion terminates
if (N == 0 )
return ;
// Print the current value of N
System.out.println(N + " " );
// Call itself recursively
Geek(N - 1 );
} // Driver code public static void main(String[] args)
{ // Initial value of N
int N = 5 ;
// Call the recursive function
Geek(N);
} } // This code is contributed by abhinavjain194 |
# Python program to demsonstrate Finite Recursion # Recursive function def Geek( N):
# Base condition
# When this condition is met,
# the recursion terminates
if (N = = 0 ):
return
# Pr the current value of N
print ( N, end = " " )
# Call itself recursively
Geek(N - 1 )
# Driver code # Initial value of N N = 5
# Call the recursive function Geek(N) # this code is contributed by shivanisinghss2110 |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Recursive function static void Geek( int N)
{ // Base condition
// When this condition is met,
// the recursion terminates
if (N == 0)
return ;
// Print the current value of N
Console.Write(N + " " );
// Call itself recursively
Geek(N - 1);
} // Driver Code public static void Main(String[] args)
{ // Initial value of N
int N = 5;
// Call the recursive function
Geek(N);
} } // This code is contributed by target_2. |
<script> // JavaScript program to demsonstrate Finite Recursion // Recursive function function Geek(N)
{ // Base condition
// When this condition is met,
// the recursion terminates
if (N == 0)
return ;
// Print the current value of N
document.write(N + " " );
// Call itself recursively
Geek(N - 1);
} // Driver code // Initial value of N var N = 5;
// Call the recursive function
Geek(N);
// this code is contributed by shivanisinghss2110
</script> |
5 4 3 2 1
Time Complexity: O(n)
Auxiliary Space: O(n)
The recursion tree for the above recursive function looks like this.
When the value of N becomes 0, because of the base condition, the recursion terminates.
Infinite Recursion:
Infinite Recursion occurs when the recursion does not terminate after a finite number of recursive calls. As the base condition is never met, the recursion carries on infinitely.
Example:
Below is an implementation to demonstrate Infinite Recursion.
// C++ program to demsonstrate Infinite Recursion #include <bits/stdc++.h> using namespace std;
// Recursive function void Geek( int N)
{ // Base condition
// This condition is never met here
if (N == 0)
return ;
// Print the current value of N
cout << N << " " ;
// Call itself recursively
Geek(N);
} // Driver code int main()
{ // Initial value of N
int N = 5;
// Call the recursive function
Geek(N);
return 0;
} |
// Java program to demsonstrate Infinite Recursion import java.io.*;
class GFG
{ // Recursive function static void Geek( int N)
{ // Base condition
// This condition is never met here
if (N == 0 )
return ;
// Print the current value of N
System.out.print( N + " " );
// Call itself recursively
Geek(N);
} // Driver code public static void main(String[] args)
{
// Initial value of N
int N = 5 ;
// Call the recursive function
Geek(N);
}
} // This code is contributed by shivanisinghss2110 |
# Python3 to demsonstrate Infinite Recursion # Recursive function def Geek(N):
# Base condition
# This condition is never met here
if (N = = 0 ):
return
# Print the current value of N
print (N, end = " " )
# Call itself recursively
Geek(N)
# Driver code # Initial value of N N = 5
# Call the recursive function Geek(N) # This code is contributed by shivanisinghss2110 |
// C# program to demsonstrate Infinite Recursion using System;
class GFG
{ // Recursive function static void Geek( int N)
{ // Base condition
// This condition is never met here
if (N == 0)
return ;
// Print the current value of N
Console.Write( N + " " );
// Call itself recursively
Geek(N);
} // Driver code public static void Main(String[] args)
{
// Initial value of N
int N = 5;
// Call the recursive function
Geek(N);
}
} // This code is contributed by shivanisinghss2110 |
<script> // JavaScript program to demsonstrate Infinite Recursion // Recursive function function Geek(N)
{ // Base condition
// This condition is never met here
if (N == 0)
return ;
// Print the current value of N
document.write( N + " " );
// Call itself recursively
Geek(N);
} // Driver code // Initial value of N
var N = 5;
// Call the recursive function
Geek(N);
// This code is contributed by shivanisinghss2110 </script> |
Time Complexity: non finite as this recursion will never end.
Auxiliary Space: non finite
The recursion tree for the above recursive function looks like this.
Since the value of N never becomes 0, so the recursion never terminates. Instead, the recursion continues until the implicit stack becomes full which results in a Stack Overflow. Some compilers directly give the output as Segmentation Fault (Core Dumped), while others may abnormally terminate for some value and then show Segmentation fault.