javascript - 如何以堆栈安全的方式映射树？

有多种递归映射树的方法：

const reduceTree = (f, node) => {
  const go = ([x, xs]) => f(x, xs.map(go));
  return go(node);
};

const mapTree = (f, node) =>
  reduceTree((x, xs) => Node_(f(x), xs), node);

const Node_ = (x, xs) => ([x, xs]);
const Node = (x, ...xs) => ([x, xs]);

const tree = Node(1,
  Node(2,
    Node(3),
    Node(4,
      Node(5))),
  Node(6),
  Node(7,
    Node(8),
    Node(9),
    Node(10),
    Node(11)));

console.log(
  mapTree(x => 2 * x, tree));

实际上，这是一个不错的解决方案，但是，它不是堆栈安全的。由于递归调用（xs.map(go)）不在尾部位置，因此我不能退回到蹦床，而且以尾部递归形式转换算法对我来说似乎也不是一件容易的事。

这样做的通常方法可能是CPS转换，这种转换会混淆计算。也许还有另一种实现堆栈安全的方法-例如使用发电机？是否存在可以以几乎机械方式应用的这种转换的一般规则？

我主要对进行此转换的过程感兴趣，而不对最终结果感兴趣。

最佳答案

从格式良好的相互递归对开始-

// map :: (a -> b, Tree a) -> Tree b
const map = (f, [ v, children ]) =>
  Node (f (v), ...mapAll (f, children))

// mapAll :: (a -> b, [ Tree a ]) -> [ Tree b ]
const mapAll = (f, nodes = []) =>
  nodes .map (n => map (f, n))

我们首先删除渴望的Array.prototype.map，它可能不允许尾部形式-

const map = (f, [ v, children ]) =>
  Node (f (v), ...mapAll (f, children))

const mapAll = (f, [ node, ...more ]) =>
  node === undefined
    ? []
    : [ map (f, node), ...mapAll (f, more) ]

接下来添加用于CPS转换的参数-

const identity = x =>
  x

const map = (f, [ v, children ], k = identity) =>
  mapAll (f, children, newChilds =>        // tail
    k (Node (f (v), ...newChilds)))        // tail

const mapAll = (f, [ node, ...more ], k = identity) =>
  node === undefined
    ? k ([])                               // tail
    : map (f, node, newNode =>             // tail
        mapAll (f, more, moreChilds =>     // tail
          k ([ newNode, ...moreChilds ]))) // tail

在下面的您自己的浏览器中验证结果

const Node = (x, ...xs) =>
  [ x, xs ]

const identity = x =>
  x

const map = (f, [ v, children ], k = identity) =>
  mapAll (f, children, newChilds =>
    k (Node (f (v), ...newChilds)))

const mapAll  = (f, [ node, ...more ], k = identity) =>
  node === undefined
    ? k ([])
    : map (f, node, newNode =>
        mapAll (f, more, moreChilds =>
          k ([ newNode, ...moreChilds ])))

const tree =
  Node
    ( 1
    , Node
        ( 2
        , Node (3)
        , Node
            ( 4
            , Node (5)
            )
        )
    , Node (6)
    , Node
        ( 7
        , Node (8)
        , Node (9)
        , Node (10)
        , Node (11)
        )
    )

console.log (map (x => x * 10, tree))

请注意，CPS本身并不会使程序具有堆栈安全性。这只是将代码放在您选择的蹦床上的形式。