Here in this Python Program, you will learn how you can delete a new node from the middle of the doubly linked list.
Let the node to be deleted is del.
- If the node to be deleted is a head node, then change the head pointer to next current head.
- Set next of previous to del, if previous to del exists.
- Set prev of next to del, if next to del exists.
Program to delete a new node from the middle of the doubly linked list.
# for Garbage collection import gc # A node of the doublly 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 to delete a node in a Doubly Linked List. # head_ref --> pointer to head node pointer. # dele --> pointer to node to be deleted def deleteNode(self, dele): # Base Case if self.head is None or dele is None: return # If node to be deleted is head node if self.head == dele: self.head = dele.next # Change next only if node to be deleted is NOT # the last node if dele.next is not None: dele.next.prev = dele.prev # Change prev only if node to be deleted is NOT # the first node if dele.prev is not None: dele.prev.next = dele.next # Finally, free the memory occupied by dele # Call python garbage collector gc.collect() # 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 program to test the above functions # Start with empty list dll = DoublyLinkedList() # Let us create the doubly linked list 10<->8<->4<->2 dll.push(2); dll.push(4); dll.push(8); dll.push(10); print "\n Original Linked List", dll.printList(dll.head) # delete nodes from doubly linked list dll.deleteNode(dll.head) dll.deleteNode(dll.head.next) dll.deleteNode(dll.head.next) # Modified linked list will be NULL<-8->NULL print "\n Modified Linked List", dll.printList(dll.head)
Original Linked list 10 8 4 2 Modified Linked list 8
The Time Complexity of the above prom is: O(1)
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