Haskell Custom Show实例

data Familia = Node String [Familia]

instance Show Familia where
  show (Node a b) = show a ++ "\n\t" ++ show b
raiz :: String -> Familia
raiz n = (Node n [])

juntar :: String -> String -> Familia -> Familia
juntar a b (Node c l) 
        | b == c = (Node c (l ++ [raiz a]))
        | otherwise = (Node c (juntarAux a b l))

juntarAux :: String -> String -> [Familia] -> [Familia]
juntarAux a b [] = []
juntarAux a b [(Node x l)]
        | x == b = [(juntar a b (Node x l))]
        | otherwise = [(Node x (juntarAux a b l))]
juntarAux a b ((Node x l):xs)
        | x == b = (juntar a b (Node x l)):xs
        | otherwise = (Node x l):(juntarAux a b xs)

这是我想要的工作方式,问题是,这是我的电流输出:

*Main> let f = raiz "Bob"
*Main> let g = juntar "John" "Bob" f
*Main> g
"Bob"
    ["John"
    []]

我想要的是这样打印:

Bob
    John
         Ruth
    Hank

所以,这个家庭的根是鲍勃,鲍勃的儿子是约翰和汉克,约翰有一个女儿叫露丝。

我在其他帖子中看到过一些东西,我尝试过几种方法,但这是最新的尝试:

instance Show Familia where
  show (Node a b) = show a ++ "\n\t" ++ (unlines $ map (unwords . map show) b)

这给了我以下错误:

t4.hs:14:77:
Couldn't match type ‘Familia’ with ‘[a0]’
Expected type: [[a0]]
  Actual type: [Familia]
In the second argument of ‘map’, namely ‘b’
In the second argument of ‘($)’, namely
  ‘map (unwords . map show) b’

有什么想法吗?提前感谢!:D