Python：使用 Pre 和 Inorder 构建二叉树

Question

我需要帮助来完成我的函数的递归部分。该函数应该使用我的 ListBinaryTree 类来帮助重建一棵树，因为它以字符串格式进行中序和前序遍历：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)

它在前序和中序遍历中找到重要值并将树拆分以确定树的左侧和右侧的数字。

它的输入和输出的一个例子是：

build_tree('3241657', '1234567')

.

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，我尝试执行以下操作：

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“索引超出范围”错误。

也适用于：

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

build_tree('3241657', '1234567')

返回

[3, None, None]
[4, None, None]
[2, None, None]
[6, None, None]
[7, None, None]
[5, None, None]
[1, None, None]    

谁能帮我解决递归部分？

谢谢

Answer 1

您使递归部分变得比必要的困难得多。

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

return ListBinaryTree(head, left_tree, right_tree)

您也许可以通过将空序列的检查移至函数顶部（例如if not inorder: return None）来进一步简化它，这样它只需要出现一次。