Javascript Program To Flatten A Multi-Level Linked List Depth Wise- Set 2

Last Updated : 03 Sep, 2024

We have discussed flattening of a multi-level linked list where nodes have two pointers down and next. In the previous post, we flattened the linked list level-wise. How to flatten a linked list when we always need to process the down pointer before next at every node.

Input:  
1 - 2 - 3 - 4
    |
    7 -  8 - 10 - 12
    |    |    |
    9    16   11
    |    |
    14   17 - 18 - 19 - 20
    |                    |
    15 - 23             21
         |
         24

Output:        
Linked List to be flattened to
1 - 2 - 7 - 9 - 14 - 15 - 23 - 24 - 8
 - 16 - 17 - 18 - 19 - 20 - 21 - 10 - 
11 - 12 - 3 - 4
Note: 9 appears before 8 (When we are 
at a node, we process down pointer before 
right pointer)

Source: Oracle Interview

If we take a closer look, we can notice that this problem is similar to tree to linked list conversion. We recursively flatten a linked list with the following steps:

If the node is NULL, return NULL.
Store the next node of the current node (used in step 4).
Recursively flatten down the list. While flattening, keep track of the last visited node, so that the next list can be linked after it.
Recursively flatten the next list (we get the next list from the pointer stored in step 2) and attach it after the last visited node.

Below is the implementation of the above idea.

JavaScript

// Javascript program to flatten 
// a multilevel linked list

// Node of Multi-level Linked List
class Node {
    constructor(val) {
        this.data = val;
        this.down = null;
        this.next = null;
    }
}
let last;

// Flattens a multi-level linked
// list depth wise
function flattenList(node) {
    if (node == null)
        return null;

    // To keep track of last visited 
    // node
    // (NOTE: This is )
    last = node;

    // Store next pointer
    let next = node.next;

    // If down list exists, process it 
    // first. Add down list as next of 
    // current node
    if (node.down != null)
        node.next = flattenList(node.down);

    // If next exists, add it after the next
    // of last added node
    if (next != null)
        last.next = flattenList(next);

    return node;
}

// Utility method to print a linked list
function printFlattenNodes(head) {
    let curr = head;
    while (curr != null) {
        console.log(curr.data + " ");
        curr = curr.next;
    }
}

// Utility function to create a 
// new node
function push(newData) {
    let newNode = new Node(newData);
    newNode.next = null;
    newNode.down = null;
    return newNode;
}

// Driver code
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.down = new Node(7);
head.next.down.down = new Node(9);
head.next.down.down.down = new Node(14);
head.next.down.down.down.down =
    new Node(15);
head.next.down.down.down.down.next =
    new Node(23);
head.next.down.down.down.down.next.down =
    new Node(24);
head.next.down.next =
    new Node(8);
head.next.down.next.down =
    new Node(16);
head.next.down.next.down.down =
    new Node(17);
head.next.down.next.down.down.next =
    new Node(18);
head.next.down.next.down.down.next.next =
    new Node(19);
head.next.down.next.down.down.next.next.next =
    new Node(20);
head.next.down.next.down.down.next.next.next.down =
    new Node(21);
head.next.down.next.next = new Node(10);
head.next.down.next.next.down = new Node(11);
head.next.down.next.next.next = new Node(12);
head = flattenList(head);
printFlattenNodes(head);
// This code is contributed by aashish1995

Output

Complexity Analysis:

Time complexity : O(n)
Space complexity : O(1)

Alternate implementation using the stack data structure:

JavaScript

function flattenList2(head) {
    let headcop = head;
    let save = new Stack();
    save.push(head);
    let prev = null;

    while (!save.isEmpty()) {
        let temp = save.pop();

        if (temp.next)
            save.push(temp.next);
        if (temp.down)
            save.push(temp.down);
        if (prev != null)
            prev.next = temp;

        prev = temp;
    }
    return headcop;
}
// This code is contributed by aashish1995