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.



C++

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// 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;
}

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Java














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// 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  



















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Python3














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# 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 



















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C#














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// 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 



















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Output:


3



          
          
          

          

    

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    Article Tags : 
Linked List
Recursion

           	
Practice Tags : 
Linked List
Recursion

        

         

         
 
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