我正在尝试编写一个函数来计算Python中二进制搜索树中的三种类型的节点。它将计算并返回包含0个子节点,1个子节点和2个子节点的节点总数。我注意到递归方法最适合这种而不是迭代方式。
def node_counts(self):
"""
---------------------------------------------------------
Returns the number of the three types of nodes in a BST.
Use: zero, one, two = bst.node_counts()
-------------------------------------------------------
Postconditions:
returns
zero - number of nodes with zero children (int)
one - number of nodes with one child (int)
two - number of nodes with two children (int)
----------------------------------------------------------
"""
zero, one, two = self._node_counts_aux(self._root)
return zero, one, two
def _node_counts_aux(self, node):
zero, one, two = 0, 0, 0
if node is not None:
if not node._right and not node._left:
zero = 1 # I understand that the problem is here.
if node._left and node._right:
two = 1 + self._node_counts_aux(node._left)[2] + self._node_counts_aux(node._right)[2]
if node._left or node._right:
one = 1 + self._node_counts_aux(node._left)[1] + self._node_counts_aux(node._right)[1]
return zero, one, two
"""
I am testing with this Tree:
36
/ \
/ \
6 50
/ \ / \
4 17 49 84
/ / / \
12 42 65 85
The output with this code comes to: (0, 6, 4).
"""
一栏在某种意义上是错误的,但在某种意义上也是对的。那不是我关心的问题。 我关注的是零,不计算在内。 零被设置为0所以我该如何解决这个问题?
答案 0 :(得分:2)
问题是方法_node_counts_aux()
会返回一个元组,但您尝试将1
添加到其结果中。您必须从递归调用中提取类型0,1和2的元素的计数,并使用这些值。
答案 1 :(得分:1)
您必须累积递归调用的结果。这可以使用zero, one, two = map(sum, zip(result_right, result_left))
完成,然后根据子项数添加相应的值。
请注意,我使用的是if/elif
语句,否则当节点有两个孩子时,您的代码也会输入一个孩子的下一个if
块。
def _node_counts_aux(self, node):
zero, one, two = 0, 0, 0
if node is not None:
result_right = self._node_counts_aux(node._right)
result_left = self._node_counts_aux(node._left)
zero, one, two = map(sum, zip(result_right, result_left))
if not node._right and not node._left:
zero += 1
elif node._left and node._right:
two += 1
elif node._left or node._right:
one += 1
return zero, one, two