如何使树节点对象在打印时显示其内容？

Question

调用查找方法时出现此错误。任何人都可以说如何纠正它？我无法使用在线可用文档对其进行调试。这是二叉树 class 的实现。我知道这与等价问题有关。

import deque

class Node:    
    def __init__(self, data):
        self.left = None
        self.right = None
        self.data = data

    def insert(self, data):    
        if data < self.data:
            if self.left is None:
                self.left = Node(data)
            else:
                self.left.insert(data)
        else:
            if self.right is None:
                self.right = Node(data)
            else:
                self.right.insert(data)  

    def lookup(self, data, parent=None):    
        if self.data == data:
            return self, parent
        if data < self.data:
            if self.left is None:
                return None
            else:
                return self.left.lookup(data, parent=self)
        else:
            if self.right is None:
                return None
            else:
                return self.right.lookup(data, parent=self)

    # aka bfs traversal
    def level_traversal(self):    
        root = self
        dq = deque()
        dq.append(root)
        while dq:
            root = dq.popleft()
            if root.left:
                dq.append(root.left)
            if root.right:
                dq.append(root.right)
            print (root.data)



    def delete(self, data):
        node, parent = self.lookup(data)
        if node.children_count() == 0:
            if parent.left == node:
                parent.left = None
            else:
                parent.right = None
            del node
        elif node.children_count() == 1:
            if node.left:
                n = node.left
            else:
                n = node.right
            if parent:
                if parent.left == node:
                    parent.left = n
                else:
                    parent.right = n
            del node
        else:
            # find the successor
            parent = node
            successor = node.right
            while successor.left:
                parent = successor
                successor = successor.left

            node.data = successor.data
            if parent.left == successor:
                parent.left = successor.right
            else:
                parent.right = successor.right

    def inorder(self):    
        if self.left:
            self.left.inorder()

        print (self.data)

        if self.right:
            self.right.inorder()

    def preorder(self):    
        print (self.data)
        if self.left:
            self.left.preorder()
        if self.right:
            self.right.preorder()

    def postorder(self):    
        if self.left:
            self.left.postorder()
        if self.right:
            self.right.postorder()
        print (self.data)


root = Node(8)
root.insert(3)
root.insert(10)
root.insert(1)
root.insert(6)
root.insert(4)
root.insert(7)
root.insert(14)
root.insert(13)

# look up
print (root.lookup(6))
# level traversal
root.level_traversal()
#mirror image
#root.mirror_image()
#root.delete(3)
#root.level_traversal()

# inorder
#root.inorder()
# pre order
#root.preorder()
# postorder
#root.postorder()
# size
#root.size()
#root.dfs()
#print root.height()

Answer 1

这根本不是错误。发生这种情况是因为您从查找方法中返回了一个对象元组，而这正是您将对象打印出来时的表示方式。如果你不喜欢这样，你可以覆盖 __repr__() 方法。

Answer 2

在您的 class 定义中试试这个。

二叉树节点的定义

class TreeNode:
  def __init__(self, val = 0, left = None, right = None):
    self.val = val
    self.left = left
    self.right = right

  ## print TreeNode Object
  def __repr__(self) -> str:
      return '[%s, %r, %r]' % (self.val, self.left, self.right)

如何使树节点对象在打印时显示其内容？

How can I make a tree Node object display its contents when printed?

python

class

binary-search-tree

二叉树节点的定义