递归邮政订单树遍历而不创建新节点

时间：2018-03-23 15:48:15

标签： javascript recursion tree tail-recursion postorder

我想定义一个适用于各种多路树的通用尾递归树遍历。这适用于预订和水平顺序，但我无法实现后订单遍历。这是我正在使用的多向树：

multi-way tree

所需订单：EKFBCGHIJDA

只要我不关心尾递归，订单遍历很容易：

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，以便我可以遍历任意树而无需了解其构造函数。有没有办法做到这一点，仍然保持尾递归？

2 个答案:

答案 0 :(得分：2)

<强>堆栈安全

我的第一个答案是通过编写我们自己的函数迭代器协议来解决这个问题。不可否认，我渴望分享这种方法，因为它是我过去探索过的东西。编写自己的数据结构非常有趣，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，您会感到无聊吗？

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' ]

下面包含其他读者的完整解决方案......

＆＃13;
＆＃13; 
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' ]
＆＃13;
＆＃13;
＆＃13;

相互递归

不知怎的，这是你的问题，让我能够画出我最有启发性的作品。回到树遍历的顶空中，我想出了这种伪应用和类Now和Later。

Later没有正确的尾调用，但我认为解决方案太过于干净而不能分享它

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, [])
}

相互递归演示

＆＃13;
＆＃13; 
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' ]
＆＃13;
＆＃13;
＆＃13;

答案 1 :(得分：1)

我们首先编写Node.value和Node.children，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，我们创建一个通用的Iterator类型。这个模仿了本机的可迭代行为，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，我们可以实施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 ())

我们可以看到它与tree此处

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

＆＃13;
＆＃13; 
// -- 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' ]
＆＃13;
＆＃13;
＆＃13;

与只有children和left字段

的节点类型相比，可变参数right字段使算法更复杂一些

这些迭代器的简化实现使它们更容易比较。在其他迭代器中为可变子数写入支持留给读者练习

// -- 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 ())

非常特别的LevelorderIterator，只是因为我认为你可以处理它

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 })

过工程改造？有罪。除了JavaScript原语外，你可以换掉上面的接口。在这里，我们将懒惰的流交换为一系列值得的值

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' ]
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