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Assume that the Indexing for the linked list always starts from 0.
If the position is greater than or equal to the length of the linked list, you should return the same linked list without any change.
The following images depict how the deletion has been performed.


You have been given a linked list of integers. Your task is to write a function that deletes a node from a given position, 'POS'.



The diameter of a binary tree is the length of the longest path between any two end nodes in a tree.
The number of edges between two nodes represents the length of the path between them.
Input: Consider the given binary tree:

Output: 6
Explanation:
Nodes in the diameter are highlighted. The length of the diameter, i.e., the path length, is 6.
You are given a Binary Tree. You are supposed to return the length of the diameter of the tree.
The diameter of a binary tree is the length of the longest path between any two end nodes in a tree.


1. If the list is empty, the function immediately returns None because there is no middle node to find.
2. If the list has only one node, then the only node in the list is trivially the middle node, and the function returns that node.
Given the head node of the singly linked list, return a pointer pointing to the middle of the linked list.
If there are an odd number of elements, return the middle element if there are even elements return the one which is farther from the head node.



The position given will always be less than or equal to the length of the linked list.
Assume that the Indexing for the linked list starts from 0.
Input :
‘K’ = 3, ‘VAL’ = 4
list = [1, 2, 3]
Output: [1, 2, 3, 4]

The ‘VAL’ = 4, is inserted at end of the above doubly linked list.
You are given a Doubly linked list, where every node in the linked list contains two pointers ‘next’ and ‘prev’ which point to the next node and previous node in the list respectively. All nodes have some positive integer value associated with them. Your task is to insert an integer value ‘VAL’ in the linked list at a given position ‘K’.
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