Flatten a multi-level linked list | Set 2 (Depth wise)

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

Code block

Output:

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

Alternate implementation using the stack data structure

C++

Node* flattenList2(Node* head)
{
    Node* headcop = head;
    stack<Node*> save;
    save.push(head);
    Node* prev = NULL;

    while (!save.empty()) {
        Node* temp = save.top();
        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;
}

Java

Node flattenList2(Node head)
{
    Node headcop = head;
    Stack<Node> save = new Stack<>();
    save.push(head);
    Node prev = null;

    while (!save.isEmpty()) {
        Node temp = save.peek();
        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 contributed by aashish1995

Python3

def flattenList2(head):

    headcop = head
    save = []
    save.append(head)
    prev = None

    while (len(save) != 0):
        temp = save[-1]
        save.pop()

        if (temp.next):
            save.append(temp.next)
        if (temp.down):
            save.append(temp.down)

        if (prev != None):
            prev.next = temp

        prev = temp

    return headcop

# This code is contributed by rutvik_56

C#

Node flattenList2(Node head)
{
    Node headcop = head;
    Stack<Node> save = new Stack<Node>();
    save.Push(head);
    Node prev = null;

    while (!save.Count != 0)
    {
        Node temp = save.Peek();
        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

Javascript

<script>
    function flattenList2(head)
    {
        var headcop = head;
          var save = new Stack();
        save.push(head);
        var prev = null;

        while (!save.isEmpty()) {
            var 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 contributed by aashish1995
</script>

Time Complexity : O(n)

Space Complexity : O(m)

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