Convert binary search tree into ordered linked list
Python: 100

class N:l=r=v=None
def T(p,q=None):
 if not p:return q
 r=p.r=T(p.r,q)
 if r:r.l=p
 return T(p.l,p)


Ungolfed Version

class Node:
    left = right = value = None


def transform(node, next=None):
    """
    node: head node to convert
    next: next node for the tail node to connect
    return: The converted head
    """
    if not node:
        return next
    node.right = transform(node.right, next)
    if node.right:
        node.right.left = node
    return transform(node.left, node)


Instead of returning a (head, tail) pair, in my solution, a next node is passed as argument to the transform function so that it only need to return the head node. This make the code shorter and cleaner.