430. Flatten a Multilevel Doubly Linked List

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 as shown in the example below.

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.

 

Example 1:

Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
Output: [1,2,3,7,8,11,12,9,10,4,5,6]
Explanation: The multilevel linked list in the input is shown.
After flattening the multilevel linked list it becomes:

Example 2:

Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation: The multilevel linked list in the input is shown.
After flattening the multilevel linked list it becomes:

Example 3:

Input: head = []
Output: []
Explanation: There could be empty list in the input.

 

Constraints:

• The number of Nodes will not exceed 1000.
• 1 <= Node.val <= 105

 

How the multilevel linked list is represented in test cases:

We use the multilevel linked list from Example 1 above:

 1---2---3---4---5---6--NULL
         |
         7---8---9---10--NULL
             |
             11--12--NULL

The serialization of each level is as follows:

[1,2,3,4,5,6,null]
[7,8,9,10,null]
[11,12,null]

To serialize all levels together, we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes:

[1,    2,    3, 4, 5, 6, null]
             |
[null, null, 7,    8, 9, 10, null]
                   |
[            null, 11, 12, null]

Merging the serialization of each level and removing trailing nulls we obtain:

[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]

Problem Explanation

The goal is to flatten a multilevel doubly linked list. Each node has next, prev, and child pointers. The child pointer points to another doubly linked list, which might have its own child pointers. We need to flatten this structure into a single-level doubly linked list.

Approach

1. Traverse the linked list using a pointer current.
2. Whenever we encounter a node with a child, we need to:
   • Temporarily store the next node.
   • Flatten the child list recursively.
   • Insert the flattened child list between the current node and the next node.
   • Update all the necessary pointers (next and prev).
   • Ensure the child pointer is set to null.
3. Continue this process until the entire list is flattened.

Code with Detailed Comments

var flatten = function(head) {
    if (head === null) return null;

    // Start traversing from the head node
    let current = head;
    
    // Traverse the entire linked list
    while (current !== null) {
        // If the current node has a child, we need to flatten it
        if (current.child !== null) {
            // Store the next node temporarily
            let next = current.next;
            
            // Recursively flatten the child list
            let child = flatten(current.child);

            // Insert the flattened child list
            current.next = child;
            child.prev = current;
            
            // Find the tail of the flattened child list
            while (child.next !== null) {
                child = child.next;
            }

            // Connect the tail of the flattened child list to the next node
            if (next !== null) {
                next.prev = child;
            }
            child.next = next;
            
            // Remove the child pointer
            current.child = null;
        }
        
        // Move to the next node
        current = current.next;
    }
    
    return head;
};

Step-by-Step Explanation

1. Initialization:
   • Check if head is null. If it is, return null immediately.
   • Initialize current to start from the head node.
2. Traversal:
   • Use a while loop to traverse the entire linked list.
3. Handling the child:
   • If current has a child, perform the following steps:
     • Temporarily store the next node (current.next).
     • Recursively flatten the child list by calling flatten(current.child).
     • Insert the flattened child list between the current node and the next node:
       • Set current.next to the head of the flattened child list.
       • Update the prev pointer of the flattened child list head to point to current.
     • Find the tail of the flattened child list by iterating through it until child.next is null.
     • Connect the tail of the flattened child list to the next node:
       • Set next.prev to the tail of the flattened child list (if next is not null).
       • Set the next pointer of the tail of the flattened child list to next.
     • Set current.child to null as it is now flattened and integrated into the main list.
4. Continue Traversal:
   • Move current to the next node (current.next) and repeat the process until the end of the list is reached.

Complexity Analysis

• Time Complexity: O(n), where n is the total number of nodes in the list. Each node is processed once, and the child lists are flattened and inserted in linear time.
• Space Complexity: O(d), where d is the maximum depth of the multilevel list. This space is used by the recursion stack due to the depth-first search approach.

This approach ensures that all pointers are correctly updated, maintaining the doubly linked structure while flattening the multilevel list into a single-level list.