Unlocking the Mysteries of the Squiggly List: A Tale of Flattening Multilevel Doubly Linked Lists
Intuition\nWhen reading the question to flatten a multilevel doubly linked list, a first intuition would be to use recursion or iteration to traverse the list a
# Intuition\nWhen reading the question to flatten a multilevel doubly linked list, a first intuition would be to use recursion or iteration to traverse the list and handle the child nodes.\n\n# Approach\nThe approach used in this code is a recursive traversal of the multilevel doubly linked list to flatten it.\nThe flatten function takes the head of the multilevel doubly linked list as input and returns the flattened version of the list.\n\n- Check if the head is None. If it is, return None.\n- Initialize an empty list T to store the values of the nodes in the desired order.\n- Invoke the traverse function with the head node and the T list as arguments.\n- The traverse function recursively traverses the multilevel doubly linked list and appends the values of the nodes to the result list.\n-- If the current node is None, return.\n-- Append the value of the current node to the result list.\n-- Recursively traverse the child node by calling the traverse function with the child node and result.\n-- Recursively traverse the next node by calling the traverse function with the next node and result.\n- After the traverse function completes, the T list contains the values of the nodes in the desired order.\n- Create a new node ret_head using the first value from the T list.\n- Initialize curr as the ret_head node.\n- Set the child pointer of ret_head to None.\n- Iterate over the remaining values in the T list from index 1.\n -- Create a new node new_node with the current value.\n -- Set the prev pointer of new_node to curr.\n -- Set the child pointer of new_node to None.\n -- Set the next pointer of curr to new_node.\n -- Update curr to the newly created new_node.\n- Return ret_head, which is the head of the flattened linked list.\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n $$O(n)$$ \n\n# Code\n```\n"""\n# Definition for a Node.\nclass Node:\n    def __init__(self, val, prev, next, child):\n        self.val = val\n        self.prev = prev\n        self.next = next\n        self.child = child\n"""\n\nclass Solution:\n    def flatten(self, head: \'Optional[Node]\') -> \'Optional[Node]\':\n        if head is None:\n            return \n        T = self.traverse(head,[])\n        ret_head = Node(T[0])\n        curr = ret_head\n        ret_head.child = None\n        for i in range(1,len(T)):\n            new_node = Node(T[i])\n            new_node.prev = curr\n            new_node.child = None\n            curr.next = new_node\n            curr = curr.next\n        return ret_head\n    def traverse(self,head,result):\n        if head is None:\n            return\n        result.append(head.val)\n        self.traverse(head.child,result)\n        self.traverse(head.next,result)\n        return result\n```
['Linked List', 'Python', 'Python3']