2011/7/23 Александр
Hello,
1) Do you really need a Monad instance for this?
Only for training purposes.
Then if you _must_ define Monad instances, maybe you should consider one for a data structure which makes more sense?
2) One possibility is just have it being (Node x _ _) >>= f = f x
I've already tried to do so, but i get only 1 element. Look.
I have a function fillTree that's fill this binary tree.
let a = fillTree 1 Empty
a >>= \x -> return (x * 2)
I got:
Node 2 Empty Empty
Well, yes, but that's the fault of your `f' ! ;-)
But i have tree with 100 elements.
OK, an alternative would be to adapt the approach taken for lists, which uses a right-fold. The problem that I see for this is how to combine the resulting trees from applying ">>= f" recursively on the right and left sub-trees with the left and right sub-trees in "f x". -- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com IvanMiljenovic.wordpress.com