From the comments it can be seen that the question is to list the root unordered labeled full binary trees. As explained in this article , the number of such trees with labels n (2n-3)!!, where !!is the double factorial function .
python ; , , :
class Node(object):
def __init__(self, left, right):
self.left = left
self.right = right
def __repr__(self):
return '(%s %s)' % (self.left, self.right)
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)
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
n == 4 (2*4 - 3)!! == 5!! == 1 * 3 * 5 == 15 trees:
>>> 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)))
, . n 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)
5 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)