8. Flatten a Multilevel Doubly Linked List.md

Question #8. Flatten a Multilevel Doubly Linked List

You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. 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.

Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.

Examples

• flatten([1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]) // should return [1,2,3,7,8,11,12,9,10,4,5,6]

  Explanation: The multilevel linked list in the input is as follows: After flattening the multilevel linked list it becomes:

• flatten([1,2,null,3]) // should return [1,3,2]

  Explanation: The input multilevel linked list is as follows:

    1---2---NULL
    |
    3---NULL
  

• flatten([]) // should return []

How multilevel linked list is represented in test case:

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]

Solution

/**
 ** Definition for a Node.
 * function Node(val,prev,next,child) {
 *    this.val = val;
 *    this.prev = prev;
 *    this.next = next;
 *    this.child = child;
 * };
 */

/**
 * @param {Node} head
 * @return {Node}
 */
function flatten(head) {
  //  Loop over the list
  for (
    let currentNode = head;
    currentNode !== null;
    currentNode = currentNode.next
  ) {
    //  If current node has a child:
    if (currentNode.child) {
      //  Get the tail of the child list
      let childTail = currentNode.child;
      //  Loop over the child list
      while (childTail.next !== null) {
        childTail = childTail.next;
      }
      //  Point the tail of the child list to the next node
      childTail.next = currentNode.next;
      //  Point the next node's prev pointer
      //  to the tail of the child list
      if (childTail.next) {
        childTail.next.prev = childTail;
      }
      //  Point the current node to the head of the child list
      currentNode.next = currentNode.child;
      //  Point the head of the child list's prev pointer
      //  to the current node
      currentNode.child.prev = currentNode;
      //  Remove the reference to the current node's child
      currentNode.child = null;
    }
  }
  //  Return the head of the flattened list
  return head;
}