append
将您作为参数传递给的对象添加到列表中,所以显然最终会生成嵌套列表。要连接的结果,所以要使用list.extend
(或+=
运营商):
>>> from collections import namedtuple
>>> class Node(namedtuple('Node', 'value left right')):
... # I reimplement __str__ just to avoid too much output later
... def __repr__(self):
... return 'Node(%r)' % self.value
... __str__ = __repr__
...
>>> def make_node(value, left=None, right=None):
... return Node(value, left, right)
...
>>> def _range(root, a, b):
... node_list = []
... if root.left:
... node_list.extend(_range(root.left, a, b))
... if a <= root.value <= b:
... node_list.append(root)
... if root.right:
... node_list.extend(_range(root.right, a, b))
... return node_list
...
>>> tree = make_node(5, make_node(3, make_node(2)), make_node(7, None, make_node(8, None, make_node(9))))
>>> _range(tree, 1, 3)
[Node(2), Node(3)]
>>> _range(tree, 1, 10)
[Node(2), Node(3), Node(5), Node(7), Node(8), Node(9)]
做这将是通过周围列表作为参数的其他方式:
>>> def _range(root, a, b, node_list=None):
... if node_list is None:
... # Note: do *not* use "[]" as a default!
... node_list = []
... if root.left:
... _range(root.left, a, b, node_list)
... if a <= root.value <= b:
... node_list.append(root)
... if root.right:
... _range(root.right, a, b, node_list)
... return node_list
...
>>> _range(tree, 1, 3)
[Node(2), Node(3)]
>>> _range(tree, 1, 10)
[Node(2), Node(3), Node(5), Node(7), Node(8), Node(9)]
您是否尝试使用'node_list.extend'而不是'node_list.append'? 另一个解决方案是将列表作为参数传递。 – Bakuriu
我花了一段时间才弄清楚你的树是从左到右,而不是从上到下。这很令人困惑...... ^^ – poke