Question

我有这些结构：

data Tree = Leaf Points | Branch Points [Tree]
deriving(Eq,Show)

data Design = Design Tree Tree Int
deriving(Eq,Show)

数据类型Tree是数据类型Design的一部分，而Design包含两个Trees：

 type Points = [(Int,Int,Int,Int)]

 design = Design mtree btree 180

我必须找到所有可能的路径：

numberPaths :: Design -> Int

并从Integer更改第三个Points，从design更改180：

type Points = [(Int,Int,Int,Int)]
design = Design mtree btree 180

changeInt :: Design -> Design

Answer 1

计算路径

要查找树中的路径数，我们递归地下降树以计算每个节点的子路径数，然后将它们相加。因此除了使用numberPaths作为接口之外，我们还需要另一个类型签名为Tree -> Int的辅助函数。我们称之为descend。

descend :: Tree -> Int

我们将Leaf计为1，将Branch计为其子树的路径数之和。所以descend变为：

descend :: Tree -> Int 
descend (Leaf _)            = 1 
descend (Branch _ subtress) = foldr (\tree sum -> sum + descend tree) 0 subtrees

函数numberPaths只需要调用descend来计算其两棵树。

numberPaths :: Design -> Int 
numberPaths (Design a b _) = descend a + descend b

Chang Int

changeInt的类型签名表示没有引入Int值来更改Design类型。我认为这没有意义，所以我冒昧地将其改为Int -> Design -> Design。

changeInt的实现遵循类似的模式。为了更改树，我们递归地沿着树的每个级别向下走。我们还引入了辅助函数changePoints。

changeTree :: Int -> Tree -> Tree
changeTree n (Leaf points)            = Leaf $ changePoints n points
changeTree n (Branch points subtrees) = Branch points' subtrees'
    where points'   = changePoints n points 
          subtrees' = map (changeTree n) subtrees

changePoints :: Int -> Points -> Points 
changePoints n points = map (\(x, y, _, z) -> (x, y, n, z)) points

changeInt只需要调用changeTree即可完成工作。

changeInt :: Int -> Design -> Design 
changeInt n (Design mtree btree _) = Design mtree' btree' n
    where mtree' = changeTree n mtree 
          btree' = changeTree n btree

Haskell Tree删除节点/计算所有可能的路径

1 个答案:

计算路径

Chang Int