我正在搜索一个实用的算法来枚举所有完整标记的二叉树。
完整二叉树是一棵树,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。
标记树是一棵树,其中所有树叶都有唯一标签。
示例:
*
|\
| \
* *
/| |\
/ | | \
T C D F
答案 0 :(得分:9)
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this paper中所述,带有n
标签的此类树的数量为(2n-3)!!
,其中!!
为double factorial function。
以下python程序基于参考文献中的递归证明;我认为代码很简单,它将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)))
对该问题的另一种可能的解释是,它寻找带有指定标签列表的有序有序完整二叉树的枚举。具有n片叶子的此类树的数量由Cn-1
的{{1}}给出。
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)