Question

在Graham Hutton的“在哈斯克尔编程”一书中，解决第10章练习3时出错。

当左侧和右侧树木上的叶子数量不超过1时，这些树木被认为是平衡的。当然，它本身似乎是平衡的。定义一个函数以确定树是否平衡。首先，定义一个计算树中叶子数量的函数。

data Tree = Leaf Int | Node Tree Tree

leaves :: Tree -> Int
leaves (Leaf _) = 1 leaves 
leaves (Node l r) = leaves l + leaves r

balanced :: Tree -> Bool
(-) :: Int -> Int -> Integer
balanced (Leaf _) = True
balanced (Node l r) = abs (leaves l - leaves r) <= 1
&& balanced l && balanced r

我对Haskell一无所知，但我的教授给了我这个任务。帮助

Answer 1

您的代码看起来很好，但是您有一些小错误。

leaves :: Tree -> Int
leaves (Leaf _) = 1         -- not 1 leaves
leaves (Node l r) = leaves l + leaves r

balanced :: Tree -> Bool
-- don't declare (-) here, it's already declared by the libraries!
balanced (Leaf _) = True
balanced (Node l r) = abs (leaves l - leaves r) <= 1
   && balanced l && balanced r   -- indent this more

haskell二叉树练习

1 个答案: