递归 post 顺序树遍历而不创建新节点

Question

我想定义一个广义尾递归树遍历，适用于所有类型的多路树。这适用于预购和等级订购，但我在实施 post 订单遍历时遇到了麻烦。这是我正在使用的多路树：

所需订单：EKFBCGHIJDA

只要我不关心尾递归post顺序遍历很容易：

const postOrder = ([x, xs]) => {
  xs.forEach(postOrder);
  console.log(`${x}`);
};

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

const tree = Node("a",
  Node("b",
    Node("e"),
    Node("f",
      Node("k"))),
  Node("c"),
  Node("d",
    Node("g"),
    Node("h"),
    Node("i"),
    Node("j")));

postOrder(tree);

另一方面，尾递归方法非常麻烦：

const postOrder = (p, q) => node => {
  const rec = ({[p]: x, [q]: forest}, stack) => {
      if (forest.length > 0) {
        const [node, ...forest_] = forest;
        stack.unshift(...forest_, Node(x));
        return rec(node, stack);
      }

      else {
        console.log(x);
        
        if (stack.length > 0) {
          const node = stack.shift();
          return rec(node, stack);
        }

        else return null;
      }
    };

  return rec(node, []);
};

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

const tree = Node("a",
  Node("b",
    Node("e"),
    Node("f",
      Node("k"))),
  Node("c"),
  Node("d",
    Node("g"),
    Node("h"),
    Node("i"),
    Node("j")));

postOrder(0, 1) (tree);

特别是，我想避免创建新节点，这样我就可以遍历任意树而不必了解它们的构造函数。有没有办法做到这一点并且仍然保持尾递归？

Answer 1

我们从编写 Node.value 和 Node.children 开始，它们从您的节点

获取两个值

// -- Node -----------------------------------------------

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

Node.value = ([ value, _ ]) =>
  value

Node.children = ([ _, children ]) =>
  children

接下来，我们创建一个通用 Iterator 类型。这个模仿原生的可迭代行为，只有我们的迭代器是持久的（不可变的）

// -- Empty ----------------------------------------------

const Empty =
  Symbol ()

const isEmpty = x =>
  x === Empty

// -- Iterator -------------------------------------------

const Yield = (value = Empty, it = Iterator ()) =>
  isEmpty (value)
    ? { done: true }
    : { done: false, value, next: it.next }

const Iterator = (next = Yield) =>
  ({ next })

const Generator = function* (it = Iterator ())
{
  while (it = it.next ())
    if (it.done)
      break
    else
      yield it.value
}

最后，我们可以实现PostorderIterator

const PostorderIterator = (node = Empty, backtrack = Iterator (), visit = false) =>
  Iterator (() =>
    visit
      ? Yield (node, backtrack)
      : isEmpty (node)
        ? backtrack.next ()
        : Node.children (node)
            .reduceRight ( (it, node) => PostorderIterator (node, it)
                         , PostorderIterator (node, backtrack, true)
                         )
            .next ())

我们可以在此处看到它与您的 tree 一起使用

// -- Demo ---------------------------------------------

const tree =
  Node ("a",
    Node ("b",
      Node ("e"),
      Node ("f",
        Node ("k"))),
    Node ("c"),
    Node ("d",
      Node ("g"),
      Node ("h"),
      Node ("i"),
      Node ("j")));

const postOrderValues =
  Array.from (Generator (PostorderIterator (tree)), Node.value)

console.log (postOrderValues)
// [ 'e', 'k', 'f', 'b', 'c', 'g', 'h', 'i', 'j', 'd', 'a' ]

程序演示

// -- Node ----------------------------------------------

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

// -- Empty ---------------------------------------------

const Empty =
  Symbol ()

const isEmpty = x =>
  x === Empty

// -- Iterator ------------------------------------------

const Yield = (value = Empty, it = Iterator ()) =>
  isEmpty (value)
    ? { done: true }
    : { done: false, value, next: it.next }
    
const Iterator = (next = Yield) =>
  ({ next })
  
const Generator = function* (it = Iterator ())
{
  while (it = it.next ())
    if (it.done)
      break
    else
      yield it.value
}

const PostorderIterator = (node = Empty, backtrack = Iterator (), visit = false) =>
  Iterator (() =>
    visit
      ? Yield (node, backtrack)
      : isEmpty (node)
        ? backtrack.next ()
        : Node.children (node)
            .reduceRight ( (it, node) => PostorderIterator (node, it)
                         , PostorderIterator (node, backtrack, true)
                         )
            .next ())
            
// -- Demo --------------------------------------------              

const tree =
  Node ("a",
    Node ("b",
      Node ("e"),
      Node ("f",
        Node ("k"))),
    Node ("c"),
    Node ("d",
      Node ("g"),
      Node ("h"),
      Node ("i"),
      Node ("j")));

const postOrderValues =
  Array.from (Generator (PostorderIterator (tree)), Node.value)
  
console.log (postOrderValues)
// [ 'e', 'k', 'f', 'b', 'c', 'g', 'h', 'i', 'j', 'd', 'a' ]

与只有 left 和 right 字段的节点类型相比，可变 children 字段使算法稍微复杂一些

这些迭代器的简化实现使它们更容易比较。在其他迭代器中编写对可变子项的支持留作 reader

的练习

// -- Node ---------------------------------------------

const Node = (value, left = Empty, right = Empty) =>
  ({ value, left, right })

// -- Iterators ----------------------------------------

const PreorderIterator = (node = Empty, backtrack = Iterator ()) =>
  Iterator (() =>
    isEmpty (node)
      ? backtrack.next ()
      : Yield (node,
          PreorderIterator (node.left,
            PreorderIterator (node.right, backtrack))))

const InorderIterator = (node = Empty, backtrack = Iterator (), visit = false) =>
  Iterator (() =>
    visit
      ? Yield (node, backtrack)
      : isEmpty (node)
        ? backtrack.next ()
        : InorderIterator (node.left,
            InorderIterator (node,
              InorderIterator (node.right, backtrack), true)) .next ())

const PostorderIterator = (node = Empty, backtrack = Iterator (), visit = false) =>
  Iterator (() =>
    visit
      ? Yield (node, backtrack)
      : isEmpty (node)
        ? backtrack.next ()
        : PostorderIterator (node.left,
            PostorderIterator (node.right,
              PostorderIterator (node, backtrack, true))) .next ())

而且很特别LevelorderIterator，只是因为我觉得你可以应付

const LevelorderIterator = (node = Empty, queue = Queue ()) =>
  Iterator (() =>
    isEmpty (node)
      ? queue.isEmpty ()
        ? Yield ()
        : queue.pop ((x, q) =>
            LevelorderIterator (x, q) .next ())
      : Yield (node,
          LevelorderIterator (Empty,
            queue.push (node.left) .push (node.right))))

// -- Queue ---------------------------------------------

const Queue = (front = Empty, back = Empty) => ({
  isEmpty: () =>
    isEmpty (front),
  push: x =>
    front
      ? Queue (front, Pair (x, back))
      : Queue (Pair (x, front), back),
  pop: k =>
    front ? front.right ? k (front.left, Queue (front.right, back))
                        : k (front.left, Queue (List (back) .reverse () .pair, Empty))
          : k (undefined, undefined)
})

// -- List ----------------------------------------------

const List = (pair = Empty) => ({
  pair:
    pair,
  reverse: () =>
    List (List (pair) .foldl ((acc, x) => Pair (x, acc), Empty)),
  foldl: (f, acc) =>
    {  
      while (pair)
        (acc = f (acc, pair.left), pair = pair.right)
      return acc
    }
})

// -- Pair ----------------------------------------------
const Pair = (left, right) =>
  ({ left, right })

Over-engineered？有罪的。您可以将上面的接口换成 JavaScript 原语。在这里，我们将惰性流换成一个急切的值数组

const postOrderValues = (node = Empty, backtrack = () => [], visit = false) =>
  () => visit
      ? [ node, ...backtrack () ]
      : isEmpty (node)
        ? backtrack ()
        : Node.children (node)
            .reduceRight ( (bt, node) => postOrderValues (node, bt)
                         , postOrderValues (node, backtrack, true)
                         )
            ()

postOrderValues (tree) () .map (Node.value)
// [ 'e', 'k', 'f', 'b', 'c', 'g', 'h', 'i', 'j', 'd', 'a' ]

Answer 2

stack-safe

我的第一个答案通过编写我们自己的函数式迭代器协议解决了这个问题。不可否认，我很想分享这种方法，因为这是我过去探索过的东西。编写自己的数据结构真的很有趣，它可以为您的问题提供创造性的解决方案 - 如果我先给出简单的答案，您会觉得无聊，不是吗？

const Empty =
  Symbol ()

const isEmpty = x =>
  x === Empty

const postOrderFold = (f = (a, b) => a, acc = null, node = Empty) =>
{
  const loop = (acc, [ node = Empty, ...nodes ], cont) =>
    isEmpty (node)
      ? cont (acc)
      : ???
  return loop (acc, [ node ], identity)
}

const postOrderValues = (node = Empty) =>
  postOrderFold ((acc, node) => [ ...acc, Node.value (node) ], [], node)

console.log (postOrderValues (tree))
// [ 'e', 'k', 'f', 'b', 'c', 'g', 'h', 'i', 'j', 'd', 'a' ]

下面为其他读者提供了完整的解决方案...

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

Node.value = ([ value, _ ]) =>
  value

Node.children = ([ _, children ]) =>
  children
  
const Empty =
  Symbol ()
  
const isEmpty = x =>
  x === Empty

const identity = x =>
  x

// tail recursive
const postOrderFold = (f = (a, b) => a, acc = null, node = Empty) =>
{
  const loop = (acc, [ node = Empty, ...nodes ], cont) =>
    isEmpty (node)
      ? cont (acc)
      : loop (acc, Node.children (node), nextAcc =>
          loop (f (nextAcc, node), nodes, cont))
  return loop (acc, [ node ], identity)
}

const postOrderValues = (node = Empty) =>
  postOrderFold ((acc, node) => [ ...acc, Node.value (node) ], [], node)

const tree =
  Node("a",
    Node("b",
      Node("e"),
      Node("f",
        Node("k"))),
    Node("c"),
    Node("d",
      Node("g"),
      Node("h"),
      Node("i"),
      Node("j")))
    
console.log (postOrderValues (tree))
// [ 'e', 'k', 'f', 'b', 'c', 'g', 'h', 'i', 'j', 'd', 'a' ]

互递归

不知何故，是您的问题让我 canvas 创作了我最受启发的作品。回到树遍历的顶空，我想出了这种 pseudo-applicative 总和类型 Now 和 Later.

Later 没有正确的尾调用，但我认为解决方案太简洁了，不能不分享它

const Empty =
  Symbol ()

const isEmpty = x =>
  x === Empty

const postOrderFold = (f = (a, b) => a, acc = null, node = Empty) =>
{
  const Now = node =>
    (acc, nodes) =>
      loop (f (acc, node), nodes)

  const Later = node =>
    (acc, nodes) =>
      loop (acc, [ ...Node.children (node) .map (Later), Now (node), ...nodes ])

  const loop = (acc, [ reducer = Empty, ...rest ]) =>
    isEmpty (reducer)
      ? acc
      : reducer (acc, rest)

  // return loop (acc, [ ...Node.children (node) .map (Later), Now (node) ])
  // or more simply ...
  return Later (node) (acc, [])
}

互递归演示

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

Node.value = ([ value, _ ]) =>
  value

Node.children = ([ _, children ]) =>
  children
  
const Empty =
  Symbol ()
  
const isEmpty = x =>
  x === Empty

const postOrderFold = (f = (a, b) => a, acc = null, node = Empty) =>
{
  const Now = node =>
    (acc, nodes) =>
      loop (f (acc, node), nodes)
    
  const Later = node =>
    (acc, nodes) =>
      loop (acc, [ ...Node.children (node) .map (Later), Now (node), ...nodes ])
    
  const loop = (acc, [ reducer = Empty, ...rest ]) =>
    isEmpty (reducer)
      ? acc
      : reducer (acc, rest)
  
  // return loop (acc, [ ...Node.children (node) .map (Later), Now (node) ])
  // or more simply ...
  return Later (node) (acc, [])
}

const postOrderValues = (node = Empty) =>
  postOrderFold ((acc, node) => [ ...acc, Node.value (node) ], [], node)

const tree =
  Node("a",
    Node("b",
      Node("e"),
      Node("f",
        Node("k"))),
    Node("c"),
    Node("d",
      Node("g"),
      Node("h"),
      Node("i"),
      Node("j")))
    
console.log (postOrderValues (tree))
// [ 'e', 'k', 'f', 'b', 'c', 'g', 'h', 'i', 'j', 'd', 'a' ]

递归 post 顺序树遍历而不创建新节点

Recursive post order tree traversal without creating new nodes

javascript

tree

recursion

tail-recursion

postorder