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  1. Practice
  2. Data Structures
  3. Linked Lists
  4. Reverse a doubly linked list
  5. Discussions

Reverse a doubly linked list

  • bhoeting + 4 comments

    Thought I'd share my recursive solution:

    Node* Reverse(Node* node)
{
  // If empty list, return
  if (!node) return NULL;
  
  // Otherwise, swap the next and prev
  Node *temp = node->next;
  node->next = node->prev;
  node->prev = temp;
  
  // If the prev is now NULL, the list
  // has been fully reversed
  if (!node->prev) return node;
  
  // Otherwise, keep going
  return Reverse(node->prev);
}
    • etayluz + 3 comments

      C solution with recursion:

      Node* Reverse(Node *head)
{
    if (head == NULL || head->next == NULL) {
        return head;
    }

    Node* nextNode = head->next;
    head->next = head->prev;
    Node* newHead = Reverse(nextNode);
    nextNode->next = head;
    return newHead;
}
      • Belphegor + 0 comments

        I'm not seeing how this can be correct, you aren't even fixing the previous links, so your code will pass only because the test case is not wide enough.

      • Kingu + 1 comment

        how does every node's prev gets changed here ?

        my C++ recursive solution

        Node* Reverse(Node* head)
{
    // Complete this function
    // Do not write the main method. 
    if(head==NULL){
        return head;
    }
    else if(head->next==NULL){
        head->next=head->prev;
        head->prev=NULL;
        return head;
    }
    Node *prevPtr=head->prev;
    Node *nextPtr=head->next;
    head->next=prevPtr;
    head->prev=nextPtr;
    return Reverse(nextPtr);
}
        • monika123_M + 0 comments

          well done ! very nice !!

      • sahil651996 + 0 comments

        Genius!

    • haotie + 1 comment

      python3 version:

      def Reverse(head):
    if not head:
        return head
    head.next, head.prev = head.prev, head.next
    if not head.prev:
        return head
    return Reverse(head.prev)
      • nagacharan + 1 comment

        will you explain this code?

        • JB555 + 1 comment

          I like this solution, simple and elegant.

          In python you can switch variables in a single line like shown above, so the code goes through each node at a time reversing the connections until you hit the final node.

          The last node returned would be the final one encountered, which is now the first element.

          • jsqihui + 1 comment

            I like this, the if not head is only for checking if the list is empty. and if not head.prev is the terminate condition for recursive, am I right?

            • JB555 + 0 comments

              Hi, yes, you are correct on both counts.

    • mohitaggarwal516 + 0 comments

      thanks man !! i was trying the similar solution bhoeting's but ' if (!node->prev) return node; ' was not striking to me...

      can anyone tell me after completion of program if we run it for the display, how is end of the linked list is handled???

    • Muntaha_Islam + 0 comments

      Will it really work with prev in here:

      if (!node->prev) return node;

      Shouldn't it be node->next?