如何修复 Python 二叉搜索树中的 NoneType 错误？

Question

我正在创建一个基本的二叉搜索树程序，其中节点必须是字符串，但是，我不断收到此错误： 'builtins.AttributeError: 'NoneType' object has no attribute 'addNode'

我有点困惑，因为我认为您必须将子节点声明为 None。我的代码如下（请原谅凌乱和额外的打印语句）：

class BinarySearchTree:
  #constructor with insertion value & left/right nodes as None
  def __init__(self, data):
    self.data = data
    self.left = None
    self.right = None
  
  #function to insert node
  def addNode(root, data):
    #when tree doesn't exist
    if root == None:
      return BinarySearchTree(data)
    else:
      #when node is already in tree
      if root.data == data:
        return root
      #smaller values go left
      elif data < root.data:
          root.left.addNode(data)
      #bigger values go right
      else:
          root.right.addNode(data)
    return root

  #function to find smallest node value (to help with deletion)  
  def smallestNode(root):
      node = root
      #loop goes to lowest left leaf
      while(node.left is not None):
          node = node.left
      return node

  #function to delete node
  def removeNode(root, data):
    if root == None:
      return root
    #when node to be deleted in smaller than root, go left
    if data < root.data:
      root.left = root.left.removeNode(data)
    #when node to be deleted in bigger than root, go right
    elif data > root.data:
        root.right = root.right.removeNode(data)
    ##when node to be deleted in the same as root...
    else:
        #when node has only 1 or 0 children
        if root.right == None:
            move = root.left
            root = None
            return move
        elif root.left == None:
            move = root.right
            root = None
            return move
        #when node has 2 children, copy then delete smallest node
        move = root.right.smallestNode()
        root.data = move.data
        root.right = root.right.removeNode(move.data)
    return root

  def findNode(root, data):
    #if current node is equal to data value then return the root
    if root.data == data or root == None:
      return root
    #if current node is greater than the data value then, search to the left
    elif data < root.data:
      return root.left.findNode(data)
    #if current node is less than the data value then, search to the right
    else:
      return root.right.findNode(data)

def createBST(keys):
  root = BinarySearchTree(keys[0])

  for i in range(1,len(keys)):
      root.addNode(keys[i])
  return root

print('Hi! Welcome to the Binary Search Tree Builder')
print('Here are your options below:')
print('1) Build new tree')
print('2) Add a node')
print('3) Find a node')
print('4) Delete a node')
choice = int(input('What would you like to do?: '))
if choice == 1:
  nodes = list(input("Enter the strings you would like to build your tree from (separate by a space): ").split())
  print(nodes)
  tree = createBST(nodes)
  print(tree)

我想知道这个错误究竟是从哪里来的，我该如何解决？另外，如果您发现我的代码中出现任何其他问题，请告诉我！

Answer 1

您将“树节点”和“树”的概念混为一谈。你所拥有的是两者的混合物。请记住，所有成员函数的第一个参数应该是“self”。用“self”替换“root”可能会让你的问题更清楚。

因此，您可以创建一个 BinaryTree 类，该类具有一个名为 self.root 的单个成员，该成员拥有一个 BinaryTreeNode 对象，一旦您有了一个节点。节点将保存left 和right 值，每个值都将具有None 或BinaryTreeNode 对象。节点类可能不需要太多代码——只需要self.left 和self.right。

BinaryTree 类将有一个 addNode 成员知道如何遍历树以找到正确的位置，但是当您找到位置时，您只需设置 self.left = BinaryTreeNode(...)。节点不知道树，因此节点不知道如何添加新节点。这是树本身的功能。

这会让你的前进道路更加清晰吗？