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Python Program To Delete Alternate Nodes Of A Linked List

  • Last Updated : 07 Feb, 2022

Given a Singly Linked List, starting from the second node delete all alternate nodes of it. For example, if the given linked list is 1->2->3->4->5 then your function should convert it to 1->3->5, and if the given linked list is 1->2->3->4 then convert it to 1->3.

Method 1 (Iterative): 
Keep track of previous of the node to be deleted. First, change the next link of the previous node and iteratively move to the next node.

Python3




# Python3 program to remove alternate
# nodes of a linked list
import math
 
# A linked list node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
         
# Deletes alternate nodes
# of a list starting with head
def deleteAlt(head):
    if (head == None):
        return
 
    # Initialize prev and node to
    # be deleted
    prev = head
    now = head.next
 
    while (prev != None and
           now != None):
         
        # Change next link of previous
        # node
        prev.next = now.next
 
        # Free memory
        now = None
 
        # Update prev and node
        prev = prev.next
        if (prev != None):
            now = prev.next
     
# UTILITY FUNCTIONS TO TEST
# fun1() and fun2()
# Given a reference (pointer to pointer)
# to the head of a list and an , push a
# new node on the front of the list.
def push(head_ref, new_data):
     
    # Allocate node
    new_node = Node(new_data)
 
    # Put in the data
    new_node.data = new_data
 
    # Link the old list off the
    # new node
    new_node.next = head_ref
 
    # Move the head to po to the
    # new node
    head_ref = new_node
    return head_ref
 
# Function to print nodes in a
# given linked list
def printList(node):
    while (node != None):
        print(node.data, end = " ")
        node = node.next
     
# Driver code
if __name__=='__main__':
     
    # Start with the empty list
    head = None
 
    # Using head=push() to construct
    # list 1.2.3.4.5
    head = push(head, 5)
    head = push(head, 4)
    head = push(head, 3)
    head = push(head, 2)
    head = push(head, 1)
 
    print("List before calling deleteAlt() ")
    printList(head)
 
    deleteAlt(head)
 
    print("List after calling deleteAlt() ")
    printList(head)
# This code is contributed by Srathore

Output: 

List before calling deleteAlt() 
1 2 3 4 5 
List after calling deleteAlt() 
1 3 5 

Time Complexity: O(n) where n is the number of nodes in the given Linked List.

Method 2 (Recursive): 
Recursive code uses the same approach as method 1. The recursive code is simple and short but causes O(n) recursive function calls for a linked list of size n.

Python3




# Deletes alternate nodes of a list
# starting with head
def deleteAlt(head):
    if (head == None):
        return
 
    node = head.next
 
    if (node == None):
        return
 
    # Change the next link of head
    head.next = node.next
 
    # Free memory allocated for node
    free(node)
 
    # Recursively call for the new
    # next of head
    deleteAlt(head.next)
# This code is contributed by Srathore

Time Complexity: O(n)

Please refer complete article on Delete alternate nodes of a Linked List for more details!
 


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