我需要帮助完成我函数的递归部分。这个函数应该使用我的ListBinaryTree类来帮助重建一个树,因为它的序和前序遍历是字符串格式的:例如
preorder = '1234567'
inorder = '3241657'
def build_tree(inorder, preorder):
head = preorder[0]
print(head)
head_pos = inorder.index(head)
print(head_pos)
left_in = inorder[:head_pos]
print(left_in)
right_in = inorder[(head_pos+1):]
print(right_in)
left_pre = preorder[1:-len(right_in)]
print(left_pre)
right_pre = preorder[-len(right_in):]
print(right_pre)
它在前序遍历和序遍历中查找重要值,并将树拆分以确定树的左侧和右侧的数字。在
其输入和输出示例如下:
^{pr2}$一。在
1
3
324
657
234
567
我用来创建树的类如下:
class ListBinaryTree:
"""A binary tree class with nodes as lists."""
DATA = 0 # just some constants for readability
LEFT = 1
RIGHT = 2
def __init__(self, root_value, left=None, right=None):
"""Create a binary tree with a given root value
left, right the left, right subtrees
"""
self.node = [root_value, left, right]
def create_tree(self, a_list):
return ListBinaryTree(a_list[0], a_list[1], a_list[2])
def insert_value_left(self, value):
"""Inserts value to the left of this node.
Pushes any existing left subtree down as the left child of the new node.
"""
self.node[self.LEFT] = ListBinaryTree(value, self.node[self.LEFT], None)
def insert_value_right(self, value):
"""Inserts value to the right of this node.
Pushes any existing left subtree down as the left child of the new node.
"""
self.node[self.RIGHT] = ListBinaryTree(value, None, self.node[self.RIGHT])
def insert_tree_left(self, tree):
"""Inserts new left subtree of current node"""
self.node[self.LEFT] = tree
def insert_tree_right(self, tree):
"""Inserts new left subtree of current node"""
self.node[self.RIGHT] = tree
def set_value(self, new_value):
"""Sets the value of the node."""
self.node[self.DATA] = new_value
def get_value(self):
"""Gets the value of the node."""
return self.node[self.DATA]
def get_left_subtree(self):
"""Gets the left subtree of the node."""
return self.node[self.LEFT]
def get_right_subtree(self):
"""Gets the right subtree of the node."""
return self.node[self.RIGHT]
def __str__(self):
return '['+str(self.node[self.DATA])+', '+str(self.node[self.LEFT])+', '+\
str(self.node[self.RIGHT])+']'
对于函数的递归部分,我尝试了如下操作:
my_tree= ListBinaryTree(head)
while my_tree.get_value() != None:
left_tree = build_tree(left_in, left_pre)
right_tree = build_tree(right_in, right_pre)
my_tree.insert_value_left(left_tree)
my_tree.insert_value_right(right_tree)
print (my_tree)
但它返回一个“索引超出范围”错误。在
同样适用于:
def build_tree(inorder, preorder):
head = preorder[0]
head_pos = inorder.index(head)
left_in = inorder[:head_pos]
right_in = inorder[(head_pos+1):]
left_pre = preorder[1:-len(right_in)]
right_pre = preorder[-len(right_in):]
if left_in:
left_tree = build_tree(left_in, left_pre)
else:
left_tree = None
if right_in:
right_tree = build_tree(right_in, right_pre)
else:
right_tree = None
my_tree = ListBinaryTree(head, left_tree, right_tree)
print(my_tree)
输入
^{pr2}$退货
[3, None, None]
[4, None, None]
[2, None, None]
[6, None, None]
[7, None, None]
[5, None, None]
[1, None, None]
谁能帮我做递归部分吗?在
谢谢
你让递归部分变得比必要的困难得多。在
您也许可以通过将空序列的检查移到函数的顶部(例如
if not inorder: return None
)来进一步简化它,这样它只需要出现一次。在相关问题 更多 >
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