因此给出以下内容:
def inorder(t):
if t:
inorder(t.left)
yield t.key
inorder(t.right)
x = [ n for n in inorder(r) ]
x
只包含根节点,为什么?
这是完整的代码;请注意,BST 实现是正确的,而使用生成器的
inorder()
实现却有些错误。
class STree(object):
def __init__(self, value):
self.key = value
self.left = None
self.right = None
def insert(r, node):
if node.key < r.key:
if r.left is None:
r.left = node
else:
insert(r.left, node)
else:
if r.right is None:
r.right = node
else:
insert(r.right, node)
def inorder(t):
if t:
inorder(t.left)
yield t.key
inorder(t.right)
r = STree(10)
insert(r, STree(12))
insert(r, STree(5))
insert(r, STree(99))
insert(r, STree(1))
tree = [ n for n in inorder(r) ]
print tree
inorder(t.left)
仅创建生成器对象,并不实际运行它。您实际上需要 yield
每个子生成器产生的所有值,如下所示:
def inorder(t):
if t:
yield from inorder(t.left)
yield t.key
yield from inorder(t.right)
请注意,
yield from
便捷语法仅在Python 3.3中引入,因此如果您使用旧版本,则必须显式迭代子生成器:
# Python 2
def inorder(t):
if t:
for key in inorder(t.left):
yield key
yield t.key
for key in inorder(t.right):
yield key