枚举所有完整（带标签）二叉树

Question

我正在寻找一个实用的算法来枚举所有的满标记二叉树

完整二叉树是一棵树，其中所有内部节点的阶数为3，叶子的阶数为1，根的阶数为2

标记树是指所有叶子都具有唯一标签的树

例如：

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

Answer 1

从评论中可以清楚地看出，问题是枚举有根的无序标记的全二叉树。如this paper中所述，具有n标签的此类树的数量为(2n-3)!!，其中!!为double factorial function

下面的python程序基于参考文献中的递归证明；我认为代码足够直截了当，可以作为算法的解释：

# 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)))

对这个问题的另一种可能的解释是，它试图枚举具有指定标签列表的根有序完整二叉树。这种有n片叶子的树的数量由Cn-1给出，从Catalan number sequence开始

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)