Skip to main content

Reverse a doubly linked list

A doubly linked list has two pointers in each node. Pointer to previous node and pointer to next node.

We have seen that reversing a singly linked list is not so easy. Because you do not have previous pointer. Since our DLL has previous pointers, its reversal must be easy.

I wrote a program earlier similar to SLL reversal- taking 3 nodes at a time and reversing links of two adjacent nodes. But I came across a better solution.

Just swap the previous link and next link for each node.

 NODEPTR reverse_list(NODEPTR head )
 {
    NODEPTR temp=head; NODEPTR t1;
    while(temp)
    {
       //swap links
       t1 = temp->prev;
       temp->prev = temp->next;
       temp->next = t1;
       //move to next node
       temp = temp->prev;
    }
    if(t1!=NULL)
       head = t1->prev;
    return head; 
}

If there was only one node in the list, then t1 will be NULL and head=t1->prev will crash the program. So verify that t1 is not NULL.

The function with complete program is here.


#include<stdio.h>  
#include<stdlib.h>
 struct node  
 {  
   int n;  
   struct node *next;  
   struct node *prev;  
 };  
 typedef struct node * NODEPTR;   
 NODEPTR create_node(int value)  
 {  
   NODEPTR temp = (NODEPTR) malloc(sizeof(struct node));  
   temp->prev = temp->next = NULL;  
   temp->n = value;  
   return temp;  
 } 

 NODEPTR append_node(NODEPTR tail, NODEPTR newnode)  
 {  
    if(tail!=NULL)
     tail->next = newnode;  
    newnode->prev = tail;  
    tail = newnode;  
    return tail;  
 } 

void display_nodes(NODEPTR head)  
 {  
   NODEPTR temp = head; 
   while (temp!= NULL)  
   {  
     printf("%d====>",temp->n);  
     temp = temp->next;  
   }  
 }  
 NODEPTR reverse_list(NODEPTR head )
 {
    NODEPTR temp=head; NODEPTR t1;
    while(temp)
    {
       t1 = temp->prev;
       temp->prev = temp->next;
       temp->next = t1;
       temp = temp->prev;
    }
    if(t1!=NULL)
       head = t1->prev;
    return head; 
}     
int main()  
 {  
     NODEPTR head,tail;  
     NODEPTR newnode;  
     int numnodes,i;  
     //initialize head and tail. Very important!!  
     head = tail = NULL;  
     printf("Number of nodes = ");  
     scanf("%d",&numnodes);      
     for(i = 0;i<numnodes;i++)  
     {  
       int value;  
       NODEPTR newnode;  
       printf("node value=");  
       scanf("%d",&value);  
       newnode = create_node(value);  
       tail = append_node(tail,newnode);
       if(head==NULL)  
       {  
          head = tail;  
       }  
     }     
     printf("The doubly linked list is ");  
     display_nodes(head); 
    
     head = reverse_list(head); 
     printf("The doubly linked list in reverse is ");  
     display_nodes(head);  
 }  

Comments

Popular posts from this blog

Linked list in C++

A linked list is a versatile data structure. In this structure, values are linked to one another with the help of addresses. I have written in an earlier post about how to create a linked list in C.  C++ has a library - standard template library which has list, stack, queue etc. data structures. But if you were to implement these data structures yourself in C++, how will you implement? If you just use new, delete, cout and cin, and then claim it is your c++ program, you are not conforming to OOPS concept. Remember you have to "keep it together". Keep all the functions and variables together - in a class. You have to have class called linked list in which there are methods - append, delete, display, insert, find, find_last. And there will also be a data - head. Defining node We need a structure for all these nodes. A struct can be used for this purpose, just like C. struct node { int val; struct node * next; }; Next we need to define our class. W

Swap nodes of a linked list

Qn: Write a function to swap the adjacent nodes of a singly linked list.i.e. If the list has nodes as 1,2,3,4,5,6,7,8, after swapping, the list should be 2,1,4,3,6,5,8,7 Image from: https://tekmarathon.com Though the question looks simple enough, it is tricky because you don't just swap the pointers. You need to take care of links as well. So let us try to understand how to go about it. Take two adjacent nodes p1 and p2 Let prevnode be previous node of p1 Now link prevnode to p2 Link p2 to p1 Link p1 to next node of p2 So the code will be prevnode -> next = p2; p1 -> next = p2 -> next; p2 -> next = p1; But what about the start node or head? head node does not have previous node If we swap head with second node, modified head should be sent back to caller  To take care of swapping first and second nodes, we can write p1 = head; p2 = head -> next; p1 -> next = p2 -> next; p2 -> next = p1; head = p2;  Now we are read

Binary tree deletion - non-recursive

In the previous post we have seen how to delete a node of a binary search tree using recursion. Today we will see how to delete a node of BST using a non-recursive function. Let us revisit the 3 scenarios here Deleting a node with no children just link the parent to NULL Deleting a node with one child link the parent to  non-null child of node to be deleted Deleting a node with both children select the successor of node to be deleted copy successor's value into this node delete the successor In order to start, we need a function to search for a node in binary search tree. Did you know that searching in  a BST is very fast, and is of the order O(logn). To search Start with root Repeat until value is found or node is NULL If the search value is greater than node branch to right If the search value is lesser than node branch to left.  Here is the function NODEPTR find_node (NODEPTR root,NODEPTR * parent, int delval) { NODEPTR nd = root; NODEPTR pa = root; if (root -> v