Find middle of singly linked list Recursively

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

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

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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; 
        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; 
} 

Output:

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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