Tower of Hanoi Algorithm (original) (raw)

Last Updated : 21 Jan, 2026

The **Tower of Hanoi is a classic mathematical puzzle involving three rods (**A, B, and C) and **n disks of different sizes. Initially, all disks are stacked on rod A in decreasing order of diameter - the largest disk at the bottom and the smallest at the top.
Goal is to move the entire stack to another rod (rod C) while following these rules:

Move only one disk at a time.
At each step, you can take the top disk from any rod and place it on another rod.
A disk can only be moved if it is the topmost disk of a rod.
No larger disk may be placed on top of a smaller disk.

**Examples:

**Input: n = 3
**Output:
Disk 1 moved from A to C
Disk 2 moved from A to B
Disk 1 moved from C to B
Disk 3 moved from A to C
Disk 1 moved from B to A
Disk 2 moved from B to C
Disk 1 moved from A to C

**Input: n = 4
**Output:
Disk 1 moved from A to B
Disk 2 moved from A to C
Disk 1 moved from B to C
Disk 3 moved from A to B
Disk 1 moved from C to A
Disk 2 moved from C to B
Disk 1 moved from A to B
Disk 4 moved from A to C
Disk 1 moved from B to C
Disk 2 moved from B to A
Disk 1 moved from C to A
Disk 3 moved from B to C
Disk 1 moved from A to B
Disk 2 moved from A to C
Disk 1 moved from B to C

Try It Yourself

**Illustrations:

**[Approach] Tower of Hanoi using Recursion

The idea is to use the helper node to reach the destination using recursion. Below is the pattern for this problem:

Shift 'n-1' disks from 'A' to 'B', using C.

Shift last disk from 'A' to 'C'.

Shift 'n-1' disks from 'B' to 'C', using A.

Follow the steps below to solve the problem:

Create a function towerOfHanoi where pass the n (current number of disk), fromRod, toRod, auxRod.
Make a function call for n - 1 th disk.
Then print the current the disk along with from_rod and to_rod
Again make a function call for n - 1 th disk. C++ `

#include using namespace std;

void towerOfHanoi(int n, char fromRod, char toRod, char auxRod){ if (n == 0) { return; } towerOfHanoi(n - 1, fromRod, auxRod, toRod); cout << "Disk " << n << " moved from " << fromRod << " to " << toRod << endl; towerOfHanoi(n - 1, auxRod, toRod, fromRod); }

int main(){ int n = 3;

// A, B and C are names of rods
towerOfHanoi(n, 'A', 'C', 'B');
return 0;

}

C

#include <stdio.h>

void towerOfHanoi(int n, char fromRod, char toRod, char auxRod) { if (n == 0) { return; } towerOfHanoi(n - 1, fromRod, auxRod, toRod); printf("Disk %d moved from %c to %c\n", n, fromRod, toRod); towerOfHanoi(n - 1, auxRod, toRod, fromRod); }

int main() { int n = 3;

// A, B and C are names of rods
towerOfHanoi(n, 'A', 'C', 'B');
return 0;

}

Java

class GFG { static void towerOfHanoi(int n, char fromRod, char toRod, char auxRod){ if (n == 0) { return; } towerOfHanoi(n - 1, fromRod, auxRod, toRod); System.out.println("Disk " + n + " moved from " + fromRod + " to " + toRod); towerOfHanoi(n - 1, auxRod, toRod, fromRod); } public static void main(String args[]){ int n = 3;

    // A, B and C are names of rods
    towerOfHanoi(n, 'A', 'C', 'B');
}

}

Python

def TowerOfHanoi(n, fromRod, toRod, auxRod): if n == 0: return TowerOfHanoi(n-1, fromRod, auxRod, toRod) print("Disk", n, " moved from ", fromRod, " to ", toRod) TowerOfHanoi(n-1, auxRod, toRod, fromRod)

if name == "main": n = 3

# A, C, B are the name of rods
TowerOfHanoi(n, 'A', 'C', 'B')

C#

using System; class GFG { static void towerOfHanoi(int n, char fromRod, char toRod, char auxRod){ if (n == 0) { return; } towerOfHanoi(n - 1, fromRod, auxRod, toRod); Console.WriteLine("Disk " + n + " moved from " + fromRod + " to " + toRod); towerOfHanoi(n - 1, auxRod, toRod, fromRod); } public static void Main(String[] args){ int n = 3;

    // A, B and C are names of rods
    towerOfHanoi(n, 'A', 'C', 'B');
}

}

JavaScript

function towerOfHanoi(n, fromRod, toRod, auxRod){ if (n == 0){ return; } towerOfHanoi(n - 1, fromRod, auxRod, toRod); console.log("Disk " + n + " moved from " + fromRod + " to " + toRod); towerOfHanoi(n - 1, auxRod, toRod, fromRod); }

// Driver code
var n = 3;

// A, B and C are names of rods
towerOfHanoi(n, 'A', 'C', 'B');

Output

Disk 1 moved from A to C Disk 2 moved from A to B Disk 1 moved from C to B Disk 3 moved from A to C Disk 1 moved from B to A Disk 2 moved from B to C Disk 1 moved from A to C

**Time complexity: O(2n), There are two possibilities for every disk. Therefore, 2 * 2 * 2 * . . . * 2(n times) is 2n
**Auxiliary Space: O(n), Function call stack space

**Related Articles