Find middle of singly linked list Recursively

Given a singly linked list and the task is to find the middle of the linked list. 

Examples:

Input  : 1->2->3->4->5  
Output : 3

Input  : 1->2->3->4->5->6
Output : 4

We have already discussed Iterative Solution. In this post iterative solution is discussed. Count total number of nodes in the list in recursive manner and do half of this, suppose this value is n. Then rolling back through recursion decrement n by one for each call. Return the node where n is zero. 

Implementation:

C++

// C++ program for Recursive approach to find
// middle of singly linked list
#include <iostream>
using namespace std;

// Tree Node Structure
struct Node
{
    int data;
    struct Node* next;
};

// Create new Node
Node* newLNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->next = NULL;
    return temp;
}

// Function for finding midpoint recursively
void midpoint_util(Node* head, int* n, Node** mid)
{

    // If we reached end of linked list
    if (head == NULL)
    {
        *n = (*n) / 2;
        return;
    }

    *n = *n + 1;

    midpoint_util(head->next, n, mid);

    // Rolling back, decrement n by one
    *n = *n - 1;
    if (*n == 0)
    {

        // Final answer
        *mid = head;
    }
}

Node* midpoint(Node* head)
{
    Node* mid = NULL;
    int n = 1;
    midpoint_util(head, &n, &mid);
    return mid;
}

int main()
{
    Node* head = newLNode(1);
    head->next = newLNode(2);
    head->next->next = newLNode(3);
    head->next->next->next = newLNode(4);
    head->next->next->next->next = newLNode(5);
    Node* result = midpoint(head);
    cout << result->data << endl;
    return 0;
}

Java

// Java program for Recursive approach to find
// middle of singly linked list
class GFG
{

// Tree Node Structure
static class Node
{
    int data;
    Node next;
};

// Create new Node
static Node newLNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.next = null;
    return temp;
}

static int n;
static Node mid;

// Function for finding midpoint recursively
static void midpoint_util(Node head )
{

    // If we reached end of linked list
    if (head == null)
    {
        n = (n) / 2;
        return;
    }

    n = n + 1;

    midpoint_util(head.next);

    // Rolling back, decrement n by one
    n = n - 1;
    if (n == 0)
    {

        // Final answer
        mid = head;
    }
}

static Node midpoint(Node head)
{
    mid = null;
    n = 1;
    midpoint_util(head);
    return mid;
}

// Driver code
public static void main(String args[])
{
    Node head = newLNode(1);
    head.next = newLNode(2);
    head.next.next = newLNode(3);
    head.next.next.next = newLNode(4);
    head.next.next.next.next = newLNode(5);
    Node result = midpoint(head);
    System.out.print( result.data );
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program for Recursive approach
# to find middle of singly linked list

# Node class
class Node:

    # Function to initialise the node object
    def __init__(self, data):
        self.data = data
        self.next = None

# Create new Node
def newLNode(data):

    temp = Node(data)
    temp.data = data
    temp.next = None
    return temp

mid = None
n = 0

# Function for finding midpoint recursively
def midpoint_util(head ):

    global n
    global mid

    # If we reached end of linked list
    if (head == None):

        n = int((n) / 2)
        return

    n = n + 1

    midpoint_util(head.next)

    # Rolling back, decrement n by one
    n = n - 1
    if (n == 0):

        # Final answer
        mid = head

def midpoint(head):

    global n
    global mid

    mid = None
    n = 1
    midpoint_util(head)
    return mid

# Driver Code
if __name__=='__main__':

    head = newLNode(1)
    head.next = newLNode(2)
    head.next.next = newLNode(3)
    head.next.next.next = newLNode(4)
    head.next.next.next.next = newLNode(5)
    result = midpoint(head)
    print( result.data )

# This code is contributed by Arnab Kundu

C#

// C# program for Recursive approach to find
// middle of singly linked list
using System;
class GFG
{

// Tree Node Structure
public class Node
{
    public int data;
    public Node next;
};

// Create new Node
static Node newLNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.next = null;
    return temp;
}

static int n;
static Node mid;

// Function for finding midpoint recursively
static void midpoint_util(Node head )
{

    // If we reached end of linked list
    if (head == null)
    {
        n = (n) / 2;
        return;
    }

    n = n + 1;

    midpoint_util(head.next);

    // Rolling back, decrement n by one
    n = n - 1;
    if (n == 0)
    {

        // Final answer
        mid = head;
    }
}

static Node midpoint(Node head)
{
    mid = null;
    n = 1;
    midpoint_util(head);
    return mid;
}

// Driver code
public static void Main()
{
    Node head = newLNode(1);
    head.next = newLNode(2);
    head.next.next = newLNode(3);
    head.next.next.next = newLNode(4);
    head.next.next.next.next = newLNode(5);
    Node result = midpoint(head);
    Console.WriteLine( result.data );
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program for Recursive approach to find
// middle of singly linked list

// Tree Node Structure
class Node
{
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
};

// Create new Node
function newLNode(data)
{
    var temp = new Node();
    temp.data = data;
    temp.next = null;
    return temp;
}

var n = 0;
var mid = null;;

// Function for finding midpoint recursively
function midpoint_util(head)
{

    // If we reached end of linked list
    if (head == null)
    {
        n = (n) / 2;
        return;
    }

    n = n + 1;

    midpoint_util(head.next);

    // Rolling back, decrement n by one
    n = n - 1;
    if (n == 0)
    {

        // Final answer
        mid = head;
    }
}

function midpoint(head)
{
    mid = null;
    n = 1;
    midpoint_util(head);
    return mid;
}

// Driver code
var head = newLNode(1);
head.next = newLNode(2);
head.next.next = newLNode(3);
head.next.next.next = newLNode(4);
head.next.next.next.next = newLNode(5);
var result = midpoint(head);
document.write( result.data );

</script>

Output

3

Time Complexity: O(N) where N is the number of nodes in the Linked List.
Auxiliary Space: O(N), due to recursion call stack