The simplest way to generically traverse a tree in haskell

Not sure if it's the simplest, but:

> data T = L | B T T deriving Data
> everything (++) (const [] `extQ` (\x -> [toConstr (x::T)])) (B L (B (B L L) L))
[B,L,B,B,L,L,L]

Here ++ says how to combine the results from subterms.

const [] is the base case for subterms who are not of type T. For those of type T, instead, we apply \x -> [toConstr (x::T)].

If you have multiple tree types, you'll need to extend the query using

const [] `extQ` (handleType1) `extQ` (handleType2) `extQ` ...

This is needed to identify the types for which we want to take the constructors. If there are a lot of types, probably this can be made shorter in some way.

Note that the code above is not very efficient on large trees since using ++ in this way can lead to quadratic complexity. It would be better, performance wise, to return a Data.Map.Map Constr Int. (Even if we do need to define some Ord Constr for that)

universe from the Data.Generics.Uniplate.Data module can give you a list of all the sub-trees of the same type. So using Ilya's example:

data T = L | B T T deriving (Data, Show)

tree :: T
tree = B L (B (B L L) L)

λ> import Data.Generics.Uniplate.Data
λ> universe tree
[B L (B (B L L) L),L,B (B L L) L,B L L,L,L,L]
λ> fmap toConstr $ universe tree
[B,L,B,B,L,L,L]