如何以堆栈安全的方式映射树？

Question

有多种方法可以递归地映射树：

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));

这实际上是一个优雅的解决方案，但它不是堆栈安全的。由于递归调用（xs.map(go)）不在尾部位置，我不能回退到蹦床，并且以尾递归形式转换算法对我来说似乎微不足道。

执行此操作的常用方法可能是CPS转换，这会使计算模糊不清。也许有另一种方法来实现堆栈安全 - 例如使用发电机？是否有这种转变的一般规则，可以几乎机械的方式应用？

我主要对进行此转换的过程感兴趣，而不是最终结果。

Answer 1

从一个结构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))

我们首先删除了不可能允许尾部形式的热切Array.prototype.map -

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CPS转换参数 -

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))

注意，CPS本身并不能使程序堆栈安全。这只是将代码放在您选择的蹦床上所需的形式。