[LeetCode] 092. Reverse Linked List II *

class Solution1:
    def reverseBetween(self, head, m, n):
        """
        :type head: ListNode
        :type m: int
        :type n: int
        :rtype: ListNode
        """

        if not head:
            return None

        left, right = head, head
        stop = False
        def recurseAndReverse(right, m, n):
            nonlocal left, stop

            # base case. Don't proceed any further
            if n == 1:
                return

            # Keep moving the right pointer one step forward until (n == 1)
            right = right.next

            # Keep moving left pointer to the right until we reach the proper node
            # from where the reversal is to start.
            if m > 1:
                left = left.next

            # Recurse with m and n reduced.
            recurseAndReverse(right, m - 1, n - 1)

            # In case both the pointers cross each other or become equal, we
            # stop i.e. don't swap data any further. We are done reversing at this
            # point.
            if left == right or right.next == left:
                stop = True

            # Until the boolean stop is false, swap data between the two pointers     
            if not stop:
                left.val, right.val = right.val, left.val

                # Move left one step to the right.
                # The right pointer moves one step back via backtracking.
                left = left.next           

        recurseAndReverse(right, m, n)
        return head