Question

我有这段代码，它在遍历后输出节点，没有任何可理解的顺序：

preorder :: Tree a -> [a]
preorder Empty = []
preorder (Branch x e d) = [x]++(preorder e)++(preorder d)

createTree = Branch 'A' 
                  (Branch 'B'
                          (Branch 'E' Empty Empty) 
                          (Branch 'B' Empty Empty)
                  )
                  (Branch 'A'
                          (Branch 'E' Empty Empty) 
                          (Branch 'A' Empty Empty)
                  )

将preorder应用于createTree输出：

ABEBAEA

我想要的是从根目录开始的所有路径的列表：

["ABE","ABB","AAE","AAA"]

我不知道如何做到这一点，我是Haskell的初学者！

非常感谢您提供的所有帮助。

Answer 1

此练习相当于：

空树的路径是什么？
让e和d成为树，每个树的路径分别列出ep和dp。如何计算标有x且e和d为子树的树的路径？

所以，

paths :: Tree a -> [[a]]
paths Empty = ???
paths (Branch x e d) = ??? -- use x,ep,dp accordingly
       where ep = paths e
             dp = paths d

作为额外提示，与您自己的示例相关：

[ 'A':xs | xs <- ["BE","BB"] ] = [ "ABE" , "ABB" ]

请注意，["BE","BB"]是第一个子树的路径。

Answer 2

paths :: Tree a -> [[a]]
paths  Empty                 = [[]]
paths (Branch x Empty d    ) = map (x:) $ paths d
paths (Branch x e     Empty) = map (x:) $ paths e
paths (Branch x e     d    ) = map (x:) $ paths e ++ paths d

即。只需在每个子路径前加x。

createTree = Branch 'A' 
                  (Branch 'B'
                          (Branch 'E' Empty (Branch 'A' Empty Empty)) 
                          (Branch 'B' Empty Empty)
                  )
                  (Branch 'A'
                          (Branch 'E' Empty Empty) 
                          (Branch 'A' Empty Empty)
                  )

main = print $ paths createTree

打印["ABEA","ABB","AAE","AAA"]

Haskell - 遍历树和输出到列表

2 个答案: