Find middle of singly linked list Recursively

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

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++ 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;

    *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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