Monad instance for binary tree

Question

I built binary tree with:

data Tree a = Empty 
              | Node a (Tree a) (Tree a)
              deriving (Eq, Ord, Read, Show)

How ca

Answer 1

There is no (good) monad for the type you just described, exactly. It would require rebalancing the tree and merging together the intermediate trees that are generated by the bind, and you can't rebalance based on any information in 'a' because you know nothing about it.

However, there is a similar tree structure

data Tree a = Tip a | Bin (Tree a) (Tree a)

which admits a monad

instance Monad Tree where
   return = Tip
   Tip a >>= f = f a
   Bin l r >>= f = Bin (l >>= f) (r >>= f)

I talked about this and other tree structures a year or two back at Boston Haskell as a lead-in to talking about finger trees. The slides there may be helpful in exploring the difference between leafy and traditional binary trees.

The reason I said there is no good monad, is that any such monad would have to put the tree into a canonical form for a given number of entries to pass the monad laws or quotient out some balance concerns by not exposing the constructors to the end user, but doing the former would require much more stringent reordering than you get for instance from an AVL or weighted tree.