Question

Haskell中的树有问题。有一棵树：

data Tree a b = Leaf a | Branch (b,Tree a b) (b,Tree a b)
    deriving(Eq, Show)

tree = Branch 
         ("A",Branch 
                ("C",Leaf 3)
                ("D",Branch
                      ("G",Leaf 7)
                      ("H",Leaf 6)
                )
          )
          ("B",Branch
                 ("E",Leaf 5)
                 ("F",Leaf 4)
          )

我需要定义一个函数，它返回此树中所有分支的列表，输出如下：[["A", "C"], ["A", "D", "G"],["A","D","H"],["B","E"],["B","F"]]。我所做的是错的，但不知道如何解决它：

branch:: Tree a b -> [[b]]
branch (Leaf x) = []
branch (Branch (a,right) (b,left)) = ([y] ++ branch left) ++ ([b] ++ branch right)

我得到的输出是["A","C","D","G","H","B","E","F"]

Answer 1

我认为这样的事情应该有效：

branch :: Tree a b -> [[b]]
branch (Leaf _) = [[]]
branch (Branch (a, right) (b, left)) = map (a :) (branch right) 
                                    ++ map (b :) (branch left)

如何在树中构建所有分支的列表？

1 个答案: