Question

考虑以下数据类型

data Tree a b = Branch b (Tree a b) (Tree a b) | Leaf a

我试图定义一个Show的实例（不导入任何模块或使用派生），它会像这样显示树

Main*> let a = Branch "x" (Branch "y" (Leaf 4) (Leaf 7)) (Leaf 9)
Main*> a
"x"
     "y"
          4
          7
     9

到目前为止，这是我提出的

findDepth (Leaf a) = 0
findDepth (Branch a (b) (c)) = 1 + (max (findDepth b) (findDepth c))
data Tree a b = Branch b (Tree a b) (Tree a b) | Leaf a
instance (Show a, Show b) => Show (Tree a b) where
     show (Leaf x) = show x
     show (Branch a (b) (c)) =
          show a ++ "\n" ++ s2 ++ show b ++ "\n" ++ s2 ++ show c ++ "\n" ++ s1
               where
                    d = findDepth (Branch a (b) (c))
                    s1 = addSpace (d-1)
                    s2 = addSpace d
                    addSpace n = replicate n '\t'

不幸的是，这会使深度最低的节点和最低深度节点的节点缩进。我知道findDepth函数实际上应该为leaf赋予最大值并分支最低值，但是无法找到为两个参数递归写入函数的方法。有什么建议吗？

Answer 1

实际上，不需要额外的findDepth功能 - 每次展示孩子时，您都可以轻松遍历树并增加深度：

import Text.Printf

data Tree a b = Branch b (Tree a b) (Tree a b) | Leaf a

instance (Show a, Show b) => Show (Tree a b) where
  show = showAtLevel 0
    where
      showAtLevel l (Leaf x) = addSpace l ++ show x
      showAtLevel l (Branch x (lt) (rt)) = printf "%s%s\n%s\n%s" (addSpace l) (show x) (showAtLevel (l + 1) lt) (showAtLevel (l + 1) rt)
      addSpace = flip replicate '\t'

测试用例：

*Main>  let a = Branch "x" (Branch "y" (Leaf 4) (Leaf 7)) (Leaf 9)
*Main> a
"x"
    "y"
        4
        7
    9
*Main> Branch "x" (Branch "y" (Leaf 4) (Branch "z" (Leaf 42) (Leaf 314))) (Leaf 9)
"x"
    "y"
        4
        "z"
            42
            314
    9

Answer 2

这里有一个没有完整解决方案的提示：编写一个单一的函数showWithDepth :: Int -> Tree -> String来传递＆＃34;累积的深度＆＃34;至今。然后你可以写show = showWithDepth 0。

请注意，一般情况下，不应该这样写show个实例，因为它的＆＃34;半标准＆＃34; show实例应该基本上像派生实例一样工作，并生成类似于有效Haskell代码的东西。（此外，在Read个实例存在的情况下，我们需要read . show === id。

显示树Haskell

2 个答案: