Recursion
Recursion is a technique where a function calls itself to solve a problem by breaking it down into smaller sub-problems.
Base Condition:
- Every function call in
recursionis stored in thecall stack. If the recursion is too deep or has no base condition, the call stack keeps growing until memory is exhausted, causing a stack overflow error. - A base condition is essential in recursion. It stops the recursion when a certain condition is met. Without it, recursion goes infinite and causes a stack overflow.
if (num === 0) return;.
Approach:
Problem:Print numbers from n to 1 usingrecursion.- Print the
number. - Recurse with
num - 1. - Stop when
num === 0.
Time Complexity: O(n)
one function call per number from n to 1.
Space Complexity: O(n)
Due to recursive call stack frames.
function printDescending(num) {
if (num === 0) return;
console.log(num);
printDescending(num - 1);
}
printDescending(5);
def print_descending(num):
if num == 0:
return
print(num)
print_descending(num - 1)
print_descending(5)
public class Main {
static void printDescending(int num) {
if (num == 0) return;
System.out.println(num);
printDescending(num - 1);
}
public static void main(String[] args) {
printDescending(5);
}
}
#include <iostream>
using namespace std;
void printDescending(int num) {
if (num == 0) return;
cout << num << endl;
printDescending(num - 1);
}
int main() {
printDescending(5);
return 0;
}
#include <stdio.h>
void printDescending(int num) {
if (num == 0) return;
printf("%d\n", num);
printDescending(num - 1);
}
int main() {
printDescending(5);
return 0;
}
using System;
class Program {
static void PrintDescending(int num) {
if (num == 0) return;
Console.WriteLine(num);
PrintDescending(num - 1);
}
static void Main() {
PrintDescending(5);
}
} 
