枚举所有完整（有标签）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。

满二叉树是所有内部节点的度数为 3、叶子的度数为 1、根的度数为 2 的树。

带标签的树是所有叶子都有唯一标签的树。

示例：

    *
    |\
    | \
    *  *
   /|  |\
  / |  | \
 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。正如本文中所解释的，带有

n

标签的此类树的数量为

(2n-3)!!

，其中

!!

是双阶乘函数。

以下Python程序基于参考论文中的递归证明；我认为代码足够简单，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)))

对该问题的另一种可能的解释是，它寻求具有指定标签列表的有根有序完整二叉树的枚举。这种具有 n 个叶子的树的数量由

Cn-1

给出，来自 加泰罗尼亚数字序列。

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)