Question

我正在学习Haskell中的数据结构。我看到了类似的东西：

  Tree(Int,Int)

这是否意味着树木的元组？我正在尝试写一些类似的东西：

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

  weight :: (Tree Integer) -> Tree(Integer,Integer)

  weight (Node left leaf right) = Node (leaf, (sum left) + (sum right)) 
                          where 
                          sum (Leaf a) = 0
                          sum (Node left leaf right) = leaf + sum left + sum right

但是我得到了无法匹配的错误。

我想得到的是每个节点的权重并将其作为元组返回，而单个叶子没有权重。

Answer 1

要修复即时错误，您需要为Node构造函数提供三个参数：

weight (Node left leaf right) = Node (weight left) 
                                     (leaf, (sum left) + (sum right)) 
                                     (weight right)
                          where 
                          sum (Leaf a) = 0
                          sum (Node left leaf right) = leaf + sum left + sum right

也许是一个基本案例

weight (Leaf a) = Leaf (a,0)

..但我不确定这是你的意图：

ghci> weight (Node (Leaf 100) 10 (Node (Leaf 20) 30 (Leaf 40)))
Node (Leaf (100,0)) (10,30) (Node (Leaf (20,0)) (30,0) (Leaf (40,0)))

忽略叶子上的值，每对中的第二个元素是子树总数。

不太经常

当你计算它们的重量时，在总结左右子树时会有很多重复。为什么不重用这个值？

我们可以通过获取顶部对中的第二个元素来读取子树的总和：

topsecond :: Tree (a,a) -> a
topsecond (Leaf (x,y)) = y
topsecond (Node _ (x,y) _) = y

所以让我们用它来得到总和

weigh (Leaf a) = Leaf (a,0)
weigh (Node left value right) = Node newleft (value,total) newright where
       newleft = weigh left
       newright = weigh right
       leftsum = topsecond newleft
       rightsum = topsecond newright
       total = leftsum + value + rightsum

以不同方式添加

在评论中你提到我们应该为标记为10的节点(10,190)。这是以下所有内容的总和，但不包括当前项目。这意味着可以通过将子树的权重与当前值相加来获得子树的总权重：

addTopPair :: Num a => Tree (a,a) -> a  -- or Tree (Integer,Integer) -> Integer
addTopPair (Leaf (x,y)) = x+y
addTopPair (Node _ (x,y) _) = x+y

然后

weighUnder (Leaf a) = Leaf (a,0)
weighUnder (Node left value right) = Node newleft (value,total) newright where
       newleft = weighUnder left
       newright = weighUnder right
       leftsum = addTopPair newleft
       rightsum = addTopPair newright
       total = leftsum + rightsum

给

ghci> weighUnder (Node (Leaf 100) 10 (Node (Leaf 20) 30 (Leaf 40)))
Node (Leaf (100,0)) (10,190) (Node (Leaf (20,0)) (30,60) (Leaf (40,0)))

和

ghci> > weighUnder $ Node (Node (Leaf (-8)) (-12) (Node (Leaf 9) 3 (Leaf 6))) 5 (Node (Leaf 2) 14 (Leaf(-2)))
Node (Node (Leaf (-8,0)) (-12,10) (Node (Leaf (9,0)) (3,15) (Leaf (6,0)))) (5,12) (Node (Leaf (2,0)) (14,0) (Leaf (-2,0)))

根据需要。

Tree（Int，Int）在haskell中的含义是什么？

1 个答案:

不太经常

以不同方式添加