430_Flatten_Multi_Level_Doubly_Linked_List

Leetcode 430: Flatten a Multilevel Doubly Linked List
Deyailed video explanation: https://youtu.be/RIyPgR7AF7M

C++:
----
class Solution {
    Node* flatten_rec(Node* head){
        Node* curr = head;
        Node* tail = head;
        while(curr){
            Node* child = curr->child;
            Node* next = curr->next;
            if(child){
                Node* _tail = flatten_rec(child);
                _tail->next = next;
                if(next) next->prev = _tail;
                curr->next = child;
                curr->child = nullptr;
                child->prev = curr;
                curr = _tail;
            }
            else
                curr = next;
            if(curr) tail = curr;
        }
        return tail;
    }
public:
    Node* flatten(Node* head) {
        if(head) flatten_rec(head);
        return head;
    }
};

Java:
----
class Solution {
    private Node flatten_rec(Node head){
        Node curr = head, tail = head;
        while(curr != null){
            Node child = curr.child;
            Node next = curr.next;
            if(child != null){
                Node _tail = flatten_rec(child);
                _tail.next = next;
                if(next != null) next.prev = _tail;
                curr.next = child;
                curr.child = null;
                child.prev = curr;
                curr = _tail;
            }
            else
                curr = next;
            if(curr != null) tail = curr;
        }
        return tail;
    }
    public Node flatten(Node head) {
        if(head != null) flatten_rec(head);
        return head;
    }
}


Python3:
-------
class Solution:
    def flatten(self, head: 'Node') -> 'Node':
        if head != None: self.flatten_rec(head)
        return head
        
    def flatten_rec(self, head):
        curr, tail = head, head
        while curr != None:
            child, next = curr.child, curr.next
            if child != None:
                _tail = self.flatten_rec(child)
                _tail.next = next
                if next != None: next.prev = _tail
                curr.next = child
                curr.child = None
                child.prev = curr
                curr = _tail
            else:
                curr = next
            if curr != None: tail = curr
        
        return tail