Reverse a Linked List - Python
Last Updated :
23 Jul, 2025
Given pointer to the head node of a linked list, the task is to reverse the linked list. We need to reverse the list by changing links between nodes.
Examples:
Input: head: 1 -> 2 -> 3 -> 4 -> NULL
Output: head: 4 -> 3 -> 2 -> 1 -> NULL
Explanation: Reversed Linked List:
Reversed Linked List
Input: head: 1 -> 2 -> 3 -> 4 -> 5 -> NULL
Output: head: 5 -> 4 -> 3 -> 2 -> 1 -> NULL
Explanation: Reversed Linked List:
Reversed Linked ListInput : NULL
Output : NULL
Input : 1->NULL
Output : 1->NULL
1. Iterative Method
The idea is to reverse the links of all nodes using three pointers:
- prev: pointer to keep track of the previous node
- curr: pointer to keep track of the current node
- next: pointer to keep track of the next node
Starting from the first node, initialize curr with the head of linked list and next with the next node of curr. Update the next pointer of curr with prev. Finally, move the three pointer by updating prev with curr and curr with next.
Python
class Node:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
# Function to initialize head
def __init__(self):
self.head = None
# Function to reverse the linked list
def reverse(self):
prev = None
current = self.head
while(current is not None):
next = current.next
current.next = prev
prev = current
current = next
self.head = prev
# Function to insert a new node at the beginning
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
# Utility function to print the LinkedList
def printList(self):
temp = self.head
while(temp):
print (temp.data,end=" ")
temp = temp.next
# Driver program to test above functions
llist = LinkedList()
llist.push(20)
llist.push(4)
llist.push(15)
llist.push(85)
print ("Given Linked List")
llist.printList()
llist.reverse()
print ("\nReversed Linked List")
llist.printList()
OutputGiven Linked List
85 15 4 20
Reversed Linked List
20 4 15 85
- Time Complexity: O(N)
- Auxiliary Space: O(1)
2. A Simpler and Tail Recursive Method
The idea is to reach the last node of the linked list using recursion then start reversing the linked list from the last node.
Python
class Node:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
# Function to initialize head
def __init__(self):
self.head = None
def reverseUtil(self, curr, prev):
# If last node mark it head
if curr.next is None:
self.head = curr
# Update next to prev node
curr.next = prev
return
# Save curr.next node for recursive call
next = curr.next
# And update next
curr.next = prev
self.reverseUtil(next, curr)
# This function mainly calls reverseUtil()
# with previous as None
def reverse(self):
if self.head is None:
return
self.reverseUtil(self.head, None)
# Function to insert a new node at the beginning
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
# Utility function to print the LinkedList
def printList(self):
temp = self.head
while(temp):
print(temp.data,end=" ")
temp = temp.next
# Driver program
llist = LinkedList()
llist.push(8)
llist.push(7)
llist.push(6)
llist.push(5)
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)
print("Given linked list")
llist.printList()
llist.reverse()
print("\nReverse linked list")
llist.printList()
OutputGiven linked list
1 2 3 4 5 6 7 8
Reverse linked list
8 7 6 5 4 3 2 1
- Time Complexity: O(N)
- Auxiliary Space: O(1)
Please refer Reverse a linked list for more details!
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