Python Program For Reversing A Doubly Linked List (original) (raw)

Last Updated : 14 Dec, 2021

Given a Doubly Linked List, the task is to reverse the given Doubly Linked List.

See below diagrams for example.

(a) Original Doubly Linked List

(b) Reversed Doubly Linked List

Here is a simple method for reversing a Doubly Linked List. All we need to do is swap prev and next pointers for all nodes, change prev of the head (or start) and change the head pointer in the end.

Python `

Program to reverse a doubly linked list

A node of the doubly linked list

class Node:

# Constructor to create a new node
def __init__(self, data):
    self.data = data
    self.next = None
    self.prev = None

class DoublyLinkedList:

# Constructor for empty Doubly 
# Linked List
def __init__(self):
    self.head = None

# Function reverse a Doubly Linked List
def reverse(self):
    temp = None
    current = self.head

    # Swap next and prev for all nodes 
    # of doubly linked list
    while current is not None:
        temp = current.prev
        current.prev = current.next
        current.next = temp
        current = current.prev

    # Before changing head, check for the
    # cases like empty list and list with 
    # only one node
    if temp is not None:
        self.head = temp.prev

# Given a reference to the head of a list 
# and an integer,inserts a new node on the 
# front of list
def push(self, new_data):

    # 1. Allocates node
    # 2. Put the data in it
    new_node = Node(new_data)

    # 3. Make next of new node as head 
    # and previous as None (already None)
    new_node.next = self.head

    # 4. change prev of head node to 
    # new_node
    if self.head is not None:
        self.head.prev = new_node

    # 5. move the head to point to the 
    # new node
    self.head = new_node

def printList(self, node):
    while(node is not None):
        print node.data,
        node = node.next

Driver code

dll = DoublyLinkedList() dll.push(2) dll.push(4) dll.push(8) dll.push(10)

print "Original Linked List" dll.printList(dll.head)

Reverse doubly linked list

dll.reverse()

print "Reversed Linked List" dll.printList(dll.head)

This code is contributed by Nikhil Kumar Singh(nickzuck_007)

`

Output:

Original linked list 10 8 4 2 The reversed Linked List is 2 4 8 10

Time Complexity: O(N), where N denotes the number of nodes in the doubly linked list.
Auxiliary Space: O(1)
We can also swap data instead of pointers to reverse the Doubly Linked List. Method used for reversing array can be used to swap data. Swapping data can be costly compared to pointers if the size of the data item(s) is more.
Please write comments if you find any of the above codes/algorithms incorrect, or find better ways to solve the same problem.

Method 2:

The same question can also be done by using Stacks.

Steps:

1. Keep pushing the node's data in the stack. -> O(n)
2. The keep popping the elements out and updating the Doubly Linked List Python3 `

""" Function to reverse a doubly-linked list swap next and prev pointers for all the nodes change prev of the head node change head pointer """ class Node: def init(self, data): self.data = data self.next = None self.prev = None

class DoublyLinkedList: def init(self): self.head = None

"""
Method to reverse a Doubly-Linked List 
using Stacks
"""
def reverseUsingStacks(self):
    stack = []
    temp = self.head
    while temp is not None:
        stack.append(temp.data)
        temp = temp.next

    # Add all the elements in the stack 
    # in a sequence to the stack
    temp = self.head
    while temp is not None:
        temp.data = stack.pop()
        temp = temp.next
        
    # Popped all the elements and the 
    # added in the linked list, 
    # in a reversed order.

"""
Method to push a new item before 
the head
"""
def push(self, new_data):
    new_node = Node(new_data)
    new_node.next = self.head

    if self.head is not None:
        self.head.prev = new_node

    self.head = new_node

"""
Method to traverse the doubly-linked 
list and print every node in the list
"""
def printList(self, node):
    while(node is not None):
        print(node.data)
        node = node. next

Driver code

dll = DoublyLinkedList() dll.push(2) dll.push(4) dll.push(8) dll.push(10)

print( "original doubly-linked list") dll.printList(dll.head)

Reverse a doubly-linked list

dll.reverseUsingStacks()

print("Reversed doubly-linked list") dll.printList(dll.head)

`

Output:

Original linked list 10 8 4 2 The reversed Linked List is 2 4 8 10

Time Complexity: O(N)
Auxiliary Space: O(N)

In this method, we traverse the linked list once and add elements to the stack, and again traverse the whole for updating all the elements. The whole takes 2n time, which is the time complexity of O(n).

Please refer complete article on Reverse a Doubly Linked List for more details!