I don't get the logic behind reversing a linked list

Question



void recursiveReverse(struct Node** head_ref)

{

    struct Node* first;

    struct Node* rest;

    /* empty list */

    if (*head_ref == NULL)

       return;  

    /* suppose first = {1, 2, 3}, rest = {2, 3} */

    first = *head_ref; 

    rest  = first->next;

    /* List has only one node */

    if (rest == NULL)     return;  

 recursiveReverse(&rest);
  first->next->next  = first; 
first->next= NULL;
*head_ref = rest;      
  }  

Answer 1

void recursiveReverse(struct Node** head_ref)
{

    struct Node* first; // Stores first element.
    struct Node* rest; // Stores rest of list.
    if (*head_ref == NULL) return;  
    first = *head_ref; 
    rest  = first->next;
    if (rest == NULL) return;  
 recursiveReverse(&rest);
// This reverses the rest part of the list.
  first->next->next  = first; 
// first->next->next is rest's last element,
// which is now linked to first to complete
// the reversal. Then to complete the list,
// we assign the next of first as NULL.
first->next= NULL;
*head_ref = rest;      
// Now the list is reversed, but head
// still points in opp order, so we assign
// it with rest.
  }  

Here is a graphical representation for 2 links:

=> 1->2->NULL
// Original List.Assume 1 as first and
// 2 as rest.

Operation : first->next->next=first;
=> 1<-2
// Last element of rear has its next
// pointing to first now.

Operation : first->next=NULL;
=> NULL<-1<-2
//Assigned first's next as null. Now
// first is the last element.

Answer 2

@Paul Victor
Can you please stop spoiling other's posts?