如何反转 Haskell 中的嵌套列表

Question

我正在尝试将类似于 (A,B,(C,(D,E)),F) 的列表反转为 return 类似于 haskell 中的 (F,((E,D),C),B,A)。我知道如何 return 一个列表：

rev :: [a] -> [a]
rev [] = []
rev (x:xs) = (rev xs) ++ [x]

但是对于嵌套列表我该如何处理呢？

Answer 1

一种可能的实现方式如下：

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 []            = []

一个简短的测试：

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，这可以更简单、更优雅地实现：

rev2 (Leaf x) = Leaf x
rev2 (Node xs) = Node (reverse (map rev2 xs))

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