Python - Delete a node at the given position in the Circular Doubly Linked List

In this method, a node at the specified position in the circular doubly linked list is deleted. For example - if the given list is 10->20->30 and the 2^nd node is deleted, the list becomes 10->20.

First, create two nodes temp and nodeToDelete to traverse through the list and track the node to delete respectively. After that count the number of elements in the list to check whether the specified position is valid or not (It must lie in the range of [1, n], where n is number of elements in the list). If the specified valid position is 1 and head is the only element in the list, then make the head as null. If the specified valid position is 1 and list contains more than one elements, then make next of head as new head and adjust links accordingly. If the specified valid position is greater than 1 then traverse to the node previous to the given position and delete the given node and adjust links accordingly.

Circular Doubly Linked List - Delete Node

The function pop_at is created for this purpose. It is a 5-step process.

def pop_at(self, position):     

  #1. create two nodes - temp and nodeToDelete 
  #   to traverse and track the node to delete  
  nodeToDelete = self.head
  temp = self.head
  NoOfElements = 0

  #2. Find the number of elements in the list
  if(temp != None):
    NoOfElements += 1
    temp = temp.next
  while(temp != self.head):
    NoOfElements += 1
    temp = temp.next

  #3. check if the specified position is valid
  if(position < 1 or position > NoOfElements):
    print("\nInavalid position.")
  elif (position == 1):
  
  #4. if the position is 1 and head is the only element
  #   in the list, then make it null, else make next
  #   of head as new head and adjust links accordingly
    if(self.head.next == self.head):
      self.head = None
    else:
      while(temp.next != self.head):
        temp = temp.next
      self.head = self.head.next
      temp.next = self.head
      self.head.prev = temp
      nodeToDelete = None

  else:

   #5. Else, traverse to the node previous to
   #   the given position and delete the given 
   #   node and adjust links accordingly 
    temp = self.head
    for i in range(1, position-1):
      temp = temp.next 
    nodeToDelete = temp.next
    temp.next = temp.next.next
    temp.next.prev = temp
    nodeToDelete = None

The below is a complete program that uses above discussed concept to delete a node at a given position in the circular doubly linked list.

# node structure
class Node:
  def __init__(self, data):
    self.data = data
    self.next = None
    self.prev = None

#class Linked List
class LinkedList:
  def __init__(self):
    self.head = None

  #Add new element at the end of the list
  def push_back(self, newElement):
    newNode = Node(newElement)
    if(self.head == None):
      self.head = newNode
      newNode.next = self.head
      newNode.prev = self.head
      return
    else:
      temp = self.head
      while(temp.next != self.head):
        temp = temp.next
      temp.next = newNode
      newNode.next = self.head
      newNode.prev = temp
      self.head.prev = newNode

  #Delete an element at the given position
  def pop_at(self, position):     
    nodeToDelete = self.head
    temp = self.head
    NoOfElements = 0

    if(temp != None):
      NoOfElements += 1
      temp = temp.next
    while(temp != self.head):
      NoOfElements += 1
      temp = temp.next

    if(position < 1 or position > NoOfElements):
      print("\nInavalid position.")
    elif (position == 1):
      if(self.head.next == self.head):
        self.head = None
      else:
        while(temp.next != self.head):
          temp = temp.next
        self.head = self.head.next
        temp.next = self.head
        self.head.prev = temp
        nodeToDelete = None

    else:
      temp = self.head
      for i in range(1, position-1):
        temp = temp.next 
      nodeToDelete = temp.next
      temp.next = temp.next.next
      temp.next.prev = temp
      nodeToDelete = None

  #display the content of the list
  def PrintList(self):
    temp = self.head
    if(temp != None):
      print("The list contains:", end=" ")
      while (True):
        print(temp.data, end=" ")
        temp = temp.next
        if(temp == self.head):
          break
      print()
    else:
      print("The list is empty.")

# test the code                  
MyList = LinkedList()

#Add three elements at the end of the list.
MyList.push_back(10)
MyList.push_back(20)
MyList.push_back(30)
MyList.PrintList()

#Delete an element at position 2
MyList.pop_at(2)
MyList.PrintList()

#Delete an element at position 1
MyList.pop_at(1)
MyList.PrintList()

The above code will give the following output:

The list contains: 10 20 30
The list contains: 10 30
The list contains: 30