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Flatten a Multilevel Doubly Linked List

Difficulty: Medium

Problem Description

You are given a doubly linked list, which contains nodes that have a next pointer, a previous pointer, and an additional child pointer. This child pointer may or may not point to a separate doubly linked list, also containing these special nodes. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure. Given the head of the first level of the list, flatten the list so that all the nodes appear in a single-level, doubly linked list. Let curr be a node with a child list. The nodes in the child list should appear after curr and before curr.next in the flattened list. Return the head of the flattened list. The nodes in the list must have all of their child pointers set to null.

Key Insights

  • The list is a multilevel doubly linked list with next, previous, and child pointers.
  • Child pointers lead to other doubly linked lists which need to be flattened and merged into the main list.
  • The solution requires depth-first traversal to ensure all nodes are visited and properly linked.
  • Maintaining proper order is essential; child nodes must be inserted between the current node and its next node.

Space and Time Complexity

Time Complexity: O(N), where N is the number of nodes in the multilevel list since we visit each node exactly once. Space Complexity: O(H), where H is the maximum depth of the multilevel list, which is used for the recursive stack space.

Solution

To solve the problem, we will use a depth-first search (DFS) approach. We will traverse the list recursively. At each node, if a child node exists, we will first flatten the child list and then connect it back to the current node. We will ensure to maintain the correct order by linking the end of the flattened child list to the next node of the current node. We will also set the child pointers of all nodes to null in the final flattened list.

Code Solutions

class Node:
    def __init__(self, val=0, prev=None, next=None, child=None):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child

def flatten(head: 'Node') -> 'Node':
    if not head:
        return head

    # Helper function to flatten the list
    def flatten_list(node):
        current = node
        last = node
        
        while current:
            if current.child:
                # Flatten the child list
                child_last = flatten_list(current.child)

                # Connect current to child
                current.child.prev = current
                current.next = current.child
                current.child = None  # Set child to None

                # Connect the end of child list to next node
                child_last.next = current.next
                if current.next:
                    current.next.prev = child_last
                
                # Move last pointer to the end of the flattened child
                last = child_last
            
            # Move to the next node
            last = current
            current = current.next
        
        return last

    flatten_list(head)
    return head
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