SDE - 2

For iteration, we create a ListNode rev to keep track of what we have reversed otherwise we would lose it. Then we iterate linked list and make head point to the current node. We change a -> to <- by calling head.next = rev, update rev by calling rev = head, move to next node by calling head = head.next. To save a temporary variable, we could assign these variables in one line, but head.next and rev should be updated before head is updated otherwise direction would not be reversed and rev would keep pointing to itself.
For example, 1->2->3, 1 is current node head, what we have reversed rev is None, 2 is head.next. Calling head.next = rev leads to None<-1. Calling head = head.next concurrently to make head pointing to 2->3. Updating rev as 1->None. And in next iteration, we will change 2->3 to 1<-2 and keep changing -> to <- so on so forth.

I didn't know the actual solution to this problem at that time.
I used a dictionary to maintain the relation between the new ListNode and the old ListNode
In a loop I create new nodes against all the old nodes in the dictionary
In the 2nd loop I created the pointers between new nodes in the dictionary.

Maintain a queue and the level at which you are currently at and then keep on adding the nodes corresponding to the level in the answers list.

Tip 1 : Learn the OOPs concepts well
Tip 2 : Focus more on how could you apply different OOPs concepts
Tip 3 : Be aware of the SOLID principles

Tip 1 : Learn the OOPs concepts well
Tip 2 : Focus more on how could you apply different OOPs concepts
Tip 3 : Be aware of the SOLID principles