Flatten a Multilevel Doubly Linked List

• Define a recursive function, let's call it flattenRec(node), which takes a node, flattens the list starting from it, and returns the tail of the flattened list.
• The main flatten function will call this recursive helper on the head and return the head.
• Inside flattenRec(node):
  1. Initialize a curr pointer to node and a tail pointer, also to node.
  2. Iterate while curr is not null.
  3. If curr has a child: a. Store curr.next in a temporary variable, nextNode. b. Make a recursive call flattenRec(curr.child) to flatten the child sublist. This call returns the tail of the flattened child list, let's call it childTail. c. Splice the child list in: set curr.next to curr.child, curr.child.prev to curr, and curr.child to null. d. Connect the end of the newly inserted child list to the rest of the original list: set childTail.next to nextNode, and if nextNode is not null, set nextNode.prev to childTail. e. Update the overall tail of the processed list to be childTail. f. Crucially, continue the iteration by setting curr to nextNode.
  4. If curr has no child, simply update tail = curr and move to the next node curr = curr.next.
• After the loop finishes, return the tail.

The core idea is to solve the problem by breaking it down into smaller, identical subproblems, which is a natural fit for recursion. We can define a recursive function that takes the head of a list (or sublist) and flattens it. A key part of the design is that this function must return the tail of the flattened list it produces. This is necessary so that the caller, which is processing a higher-level list, knows where to attach the rest of its own list.

The process for a given node curr is as follows: if curr has a child, we first recursively flatten that child's list. Once that's done and we have the tail of the flattened child list, we can 'splice' this list into the main list. This involves re-wiring the next and prev pointers of curr, the head of the child list, the tail of the child list, and curr's original next node. After splicing, we set curr.child to null as required and continue our traversal from where the spliced list ends.

1/*2// Definition for a Node.3class Node {4    public int val;5    public Node prev;6    public Node next;7    public Node child;8};9*/10class Solution {11    public Node flatten(Node head) {12        if (head == null) {13            return null;14        }15        flattenRec(head);16        return head;17    }18 19    // Flattens the list starting at 'node' and returns the tail of the flattened list.20    private Node flattenRec(Node node) {21        Node curr = node;22        Node tail = node; // Keep track of the last node visited23 24        while (curr != null) {25            Node nextNode = curr.next;26            27            if (curr.child != null) {28                // Flatten the child list and get its tail29                Node childTail = flattenRec(curr.child);30                31                // Splice the child list in32                curr.next = curr.child;33                curr.child.prev = curr;34                curr.child = null;35                36                // Connect the child tail to the original next node37                childTail.next = nextNode;38                if (nextNode != null) {39                    nextNode.prev = childTail;40                }41                42                // The new tail of the list processed so far is the child's tail43                tail = childTail;44                // The next node to process is what was originally after the child tail45                curr = nextNode; 46            } else {47                // No child, just move to the next node48                tail = curr;49                curr = nextNode;50            }51        }52        return tail;53    }54}