我正在嘗試反轉類似于(A,B,(C,(D,E)),F)回傳(F,((E,D),C),B,A)Haskell 中的內容的串列。我知道如何回傳單個串列：
rev :: [a] -> [a]
rev [] = []
rev (x:xs) = (rev xs) [x]
但是我將如何處理嵌套串列？
uj5u.com熱心網友回復：
一個可能的實作如下：
data Tree a = Leaf a | Node [Tree a] deriving (Eq, Show)
rev (Leaf x) = Leaf x
rev (Node xs) = Node (go (reverse xs)) where
go ((Leaf y):ys) = Leaf y: go ys
go ((Node y):ys) = rev (Node y): go ys
go [] = []
一個簡短的測驗：
λ> tree = Node [Leaf 'A', Leaf 'B', Node [Leaf 'C', Node [Leaf 'D', Leaf 'E']]]
λ> rev tree
Node [Node [Node [Leaf 'E',Leaf 'D'],Leaf 'C'],Leaf 'B',Leaf 'A']
正如 Daniel Wagner 所指出的，這可以更簡單、更優雅地實作：
rev2 (Leaf x) = Leaf x
rev2 (Node xs) = Node (reverse (map rev2 xs))
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