I wrote the following code. It is working and using the Reader monad.
Could you give me some hints about code style in Haskell ? Mainly, I mean monads -- I am newbie.
import Control.Monad.Reader
data Tree a = Node a (Tree a) (Tree a)
| Empty
renumberM :: Tree a -> Reader Int (Tree Int)
renumberM (Node _ l r) = ask >>= (\x ->
return (Node x (runReader (local (+1) (renumberM l)) x)
(runReader (local (+1) (renumberM r)) x)))
renumberM Empty = return Empty
renumber'' :: Tree a -> Tree Int
renumber'' t = runReader (renumberM t) 0
I want to show you that your idea is an instance of a more general concept - zipping. Here's a version of your program that employs a simpler and more functional style.
Applicative Functors
Here's the definition of Applicative:
class Functor f => Applicative f where
pure :: a -> f a
(<*>) :: f (a -> b) -> f a -> f b
You could say that a type f x is a structure f containing some values x. The function <*> takes a structure of functions (f (a -> b)) and applies it to a structure of arguments (f a) to produce a structure of results (f b).
Zippy Applicatives
One way to make Tree an applicative functor is by making <*> traverse the two trees in lock-step, zipping them together just like zip does with lists. Every time you encounter a Node in the tree of functions and a Node in the tree of arguments, you can pull the function out and apply it to the argument. You have to stop traversing when you reach the bottom of either of the trees.
instance Applicative Tree where
pure x = let t = Node x t t in t
Empty <*> _ = Empty
_ <*> Empty = Empty
(Node f lf rf) <*> (Node x lx rx) = Node (f x) (lf <*> lx) (rf <*> rx)
instance Functor Tree where
fmap f x = pure f <*> x -- as usual
pure x generates an infinite tree of xs. This works fine because Haskell is a lazy language.
+-----x-----+
| |
+--x--+ +--x--+
| | | |
+-x-+ +-x-+ +-x-+ +-x-+
| | | | | | | |
etc
So the shape of the tree t <*> pure x is the same as the shape of t: you only stop traversing when you encounter an Empty, and there aren't any in pure x. (The same applies to pure x <*> t.)
This is a common way to make a data structure an instance of Applicative. For example, the standard library includes ZipList, whose Applicative instance is very similar to that of our tree:
newtype ZipList a = ZipList { getZipList :: [a] }
instance Applicative ZipList where
pure x = ZipList (repeat x)
ZipList fs <*> ZipList xs = ZipList (zipWith ($) fs xs)
Once again, pure generates an infinite ZipList, and <*> consumes its arguments in lock-step.
The prototypical zippy Applicative, if you like, is the "reader" Applicative (->) r, which combines functions by applying them all to a fixed argument and collecting the results. So all Representable functors admit (at least) a zippy Applicative instance.
Using some Applicative machinery, we can generalise the Prelude's zip to any applicative functor (although it'll only behave precisely like zip when the Applicative is zippy in nature - for example, with the regular Applicative instance for [] zipA will give you the Cartesian product of its arguments).
zipA :: Applicative f => f a -> f b -> f (a, b)
zipA = liftA2 (,)
Labelling as Zipping
The plan is to zip the input tree together with an infinite tree containing the depth of each level. The output will be a tree with the same shape as the input tree (because the depths-tree is infinite), but each node will be labelled with its depth.
depths :: Tree Integer
depths = go 0
where go n = let t = go (n+1) in Node n t t
This is what depths looks like:
+-----0-----+
| |
+--1--+ +--1--+
| | | |
+-2-+ +-2-+ +-2-+ +-2-+
| | | | | | | |
etc
Now that we've set up the structures we need, labelling a tree is easy.
labelDepths :: Tree a -> Tree (Integer, a)
labelDepths = zipA depths
Relabelling a tree by throwing away the original labels, as you originally specified, is easy too.
relabelDepths :: Tree a -> Tree Integer
relabelDepths t = t *> depths
A quick test:
ghci> let myT = Node 'x' (Node 'y' (Node 'z' Empty Empty) (Node 'a' Empty Empty)) (Node 'b' Empty Empty)
ghci> labelDepths myT
Node (0,'x') (Node (1,'y') (Node (2,'z') Empty Empty) (Node (2,'a') Empty Empty)) (Node (1,'b') Empty Empty)
+--'x'-+ +--(0,'x')-+
| | labelDepths | |
+-'y'-+ 'b' ~~> +-(1,'y')-+ (1,'b')
| | | |
'z' 'a' (2,'z') (2,'a')
You can devise different labelling schemes by varying the tree you zip along. Here's one which tells you the path you took to reach a node:
data Step = L | R
type Path = [Step]
paths :: Tree Path
paths = go []
where go path = Node path (go (path ++ [L])) (go (path ++ [R]))
+--------[ ]--------+
| |
+---[L]---+ +---[R]---+
| | | |
+-[L,L]-+ +-[L,R]-+ +-[R,L]-+ +-[R,R]-+
| | | | | | | |
etc
(The inefficient nesting of calls to ++ above can be mitigated using difference lists.)
labelPath :: Tree a -> Tree (Path, a)
labelPath = zipA paths
As you continue to learn Haskell, you'll get better at spotting when a program is an example of a deeper concept. Setting up general structures, like I did with the Applicative instance above, quickly pays dividends in code reuse.
There's no need to go in and out of the Reader the way you do it here by using runReader; instead, you can rewrite it as
renumberR :: Tree a -> Reader Int (Tree Int)
renumberR (Node _ l r) = do
x <- ask
l' <- local (+1) (renumberR l)
r' <- local (+1) (renumberR r)
return (Node x l' r')
renumberR Empty = return Empty
However, you can write it even nicer by just using the applicative interface of Reader:
renumberR (Node _ l r) =
Node <$> ask <*> local (+1) (renumberR l) <*> local (+1) (renumberR r)
renumberR Empty = pure Empty
Note that I have renamed your function to renumberR to emphasize the fact that it runs in Reader, but not necessarily using its monadic interface.
来源：https://stackoverflow.com/questions/36474647/haskell-depth-for-each-node-in-binary-tree-using-reader-monad