在递归二叉树函数中返回元组时遇到问题
Posted
技术标签:
【中文标题】在递归二叉树函数中返回元组时遇到问题【英文标题】:Trouble returning tuple in a recursive binary tree function 【发布时间】:2022-01-07 16:14:00 【问题描述】:我正在编写一个函数,它递归地找到二叉树中的最小值和最大值并返回一个元组 (min, max)。我的代码为我的测试用例返回了正确的最小值和最大值,但分别返回。任何关于如何让它返回元组的建议都值得赞赏! (请注意,我不允许使用为我遍历树的 LinkedBinaryTree 函数,但我在我的代码下附加了我当前正在使用的类)
我的代码
from LinkedBinaryTree import LinkedBinaryTree
def min_and_max(bin_tree):
if bin_tree is None:
raise Exception("Tree is empty")
def subtree_min_and_max(root):
minval = root.data
maxval = root.data
if (root.left is None and root.right is None):
temp = (minval, maxval)
return temp
elif (root.left is None):
ltemp = subtree_min_and_max(root.right)
if (ltemp[0] < minval):
minval= ltemp[0]
if (ltemp[1] > maxval):
maxval = ltemp[1]
temp = (minval, maxval)
return temp
elif (root.right is None):
subtree_min_and_max(root.left)
rtemp = subtree_min_and_max(root.right)
if (rtemp[0] < minval):
minval = rtemp[0]
if (rtemp[1] > maxval):
maxval = rtemp[1]
temp = (minval, maxval)
return temp
else:
ltemp = subtree_min_and_max(root.left)
rtemp = subtree_min_and_max(root.right)
print((min(root.data, ltemp[0], rtemp[0])))
print(max(root.data, ltemp[1], rtemp[1]))
temp = (min(root.data, ltemp[0], rtemp[0]), max(root.data, ltemp[1], rtemp[1]))
return temp
return subtree_min_and_max(bin_tree.root)
LinkedBinaryTree 类
from ArrayQueue import ArrayQueue
class LinkedBinaryTree:
class Node:
def __init__(self, data, left=None, right=None):
self.data = data
self.parent = None
self.left = left
if (self.left is not None):
self.left.parent = self
self.right = right
if (self.right is not None):
self.right.parent = self
def __init__(self, root=None):
self.root = root
self.size = self.count_nodes()
def __len__(self):
return self.size
def is_empty(self):
return len(self) == 0
def count_nodes(self):
def subtree_count(root):
if (root is None):
return 0
else:
left_count = subtree_count(root.left)
right_count = subtree_count(root.right)
return 1 + left_count + right_count
return subtree_count(self.root)
def sum(self):
def subtree_sum(root):
if (root is None):
return 0
else:
left_sum = subtree_sum(root.left)
right_sum = subtree_sum(root.right)
return root.data + left_sum + right_sum
return subtree_sum(self.root)
def height(self):
def subtree_height(root):
if (root.left is None and root.right is None):
return 0
elif (root.left is None):
return 1 + subtree_height(root.right)
elif (root.right is None):
return 1 + subtree_height(root.left)
else:
left_height = subtree_height(root.left)
right_height = subtree_height(root.right)
return 1 + max(left_height, right_height)
if(self.is_empty()):
raise Exception("Tree is empty")
return subtree_height(self.root)
def preorder(self):
def subtree_preorder(root):
if (root is None):
pass
else:
yield root
yield from subtree_preorder(root.left)
yield from subtree_preorder(root.right)
yield from subtree_preorder(self.root)
def postorder(self):
def subtree_postorder(root):
if (root is None):
pass
else:
yield from subtree_postorder(root.left)
yield from subtree_postorder(root.right)
yield root
yield from subtree_postorder(self.root)
def inorder(self):
def subtree_inorder(root):
if (root is None):
pass
else:
yield from subtree_inorder(root.left)
yield root
yield from subtree_inorder(root.right)
yield from subtree_inorder(self.root)
def breadth_first(self):
if (self.is_empty()):
return
line = ArrayQueue()
line.enqueue(self.root)
while (line.is_empty() == False):
curr_node = line.dequeue()
yield curr_node
if (curr_node.left is not None):
line.enqueue(curr_node.left)
if (curr_node.right is not None):
line.enqueue(curr_node.right)
def __iter__(self):
for node in self.breadth_first():
yield node.data
我的测试代码
root = LinkedBinaryTree.Node(3)
T = LinkedBinaryTree(root)
a = LinkedBinaryTree.Node(2)
a.parent = root
root.left = a
b = LinkedBinaryTree.Node(7)
b.parent = root
root.right = b
c = LinkedBinaryTree.Node(9)
c.parent = a
a.left = c
d = LinkedBinaryTree.Node(5)
d.parent = c
c.left = d
e = LinkedBinaryTree.Node(1)
e.parent = c
c.right = e
f = LinkedBinaryTree.Node(8)
f.parent = b
b.left = f
g = LinkedBinaryTree.Node(4)
g.parent = b
b.right = g
print(min_and_max(T))
测试代码生成一棵看起来像这样的树
【问题讨论】:
【参考方案1】:在subtree_min_and_max
,当root.right is None
的情况下,你的代码做:
subtree_min_and_max(root.left)
rtemp = subtree_min_and_max(root.right)
应该是单身
rtemp = subtree_min_and_max(root.left)
此时,子树的右孩子为空,程序应在左孩子中搜索 min 和 max。
【讨论】:
以上是关于在递归二叉树函数中返回元组时遇到问题的主要内容,如果未能解决你的问题,请参考以下文章