python - 枚举所有完整（标记）二叉树

Question

我正在寻找一种实用的算法来枚举所有完整标记的二叉树。

完全二叉树是一棵树，其中所有内部节点的度数为 3，叶子的度数为 1，根的度数为 2。

带标签的树是所有叶子都有唯一标签的树。

例子：

    *
    |\
    | \
    *  *
   /|  |\
  / |  | \
 T  C  D  F

Answer 1

从评论中可以清楚地看出，问题是枚举有根无序标记的完整二叉树。正如本文n所解释的，这种带有标签的树的数量(2n-3)!!是!!双因子函数。

以下python程序基于参考论文中的递归证明；我认为代码很简单，可以作为算法的解释传递：

# A very simple representation for Nodes. Leaves are anything which is not a Node.
class Node(object):
  def __init__(self, left, right):
    self.left = left
    self.right = right

  def __repr__(self):
    return '(%s %s)' % (self.left, self.right)

# Given a tree and a label, yields every possible augmentation of the tree by
# adding a new node with the label as a child "above" some existing Node or Leaf.
def add_leaf(tree, label):
  yield Node(label, tree)
  if isinstance(tree, Node):
    for left in add_leaf(tree.left, label):
      yield Node(left, tree.right)
    for right in add_leaf(tree.right, label):
      yield Node(tree.left, right)

# Given a list of labels, yield each rooted, unordered full binary tree with
# the specified labels.
def enum_unordered(labels):
  if len(labels) == 1:
    yield labels[0]
  else:
    for tree in enum_unordered(labels[1:]):
      for new_tree in add_leaf(tree, labels[0]):
        yield new_tree

对于n == 4，有(2*4 - 3)!! == 5!! == 1 * 3 * 5 == 15树：

>>> for tree in enum_unordered(("a","b","c","d")): print tree
... 
(a (b (c d)))
((a b) (c d))
(b (a (c d)))
(b ((a c) d))
(b (c (a d)))
(a ((b c) d))
((a (b c)) d)
(((a b) c) d)
((b (a c)) d)
((b c) (a d))
(a (c (b d)))
((a c) (b d))
(c (a (b d)))
(c ((a b) d))
(c (b (a d)))

该问题的另一种可能解释是，它寻求具有指定标签列表的有根有序完整二叉树的枚举。具有 n 片叶子的树的数量由加泰罗尼亚数列给出。Cn-1

def enum_ordered(labels):
  if len(labels) == 1:
    yield labels[0]
  else:
    for i in range(1, len(labels)):
      for left in enum_ordered(labels[:i]):
        for right in enum_ordered(labels[i:]):
          yield Node(left, right)

对于 5 个标签，我们有：C5-1 == 14

>>> for tree in enum_ordered(("a","b","c","d", "e")): print tree
... 
(a (b (c (d e))))
(a (b ((c d) e)))
(a ((b c) (d e)))
(a ((b (c d)) e))
(a (((b c) d) e))
((a b) (c (d e)))
((a b) ((c d) e))
((a (b c)) (d e))
(((a b) c) (d e))
((a (b (c d))) e)
((a ((b c) d)) e)
(((a b) (c d)) e)
(((a (b c)) d) e)
((((a b) c) d) e)