Question

我正在尝试实现二叉搜索树，但面临一些问题在节点插入中。任何人都可以帮我指出我的python程序有什么问题吗？ addChild（）函数是不是正确添加左子（4，＆＃34; hans＆＃34;）？我的递归函数有问题吗？提前谢谢。

class Node:
    def __init__(self, key = None, value = None, leftChild = None, rightChild = None, parent = None):
        self.key = key
        self.value = value
        self.leftChild = leftChild
        self.rightChild = rightChild
        self.parent = parent

    def get_key(self):
        return self.key

    def get_value(self):
        return self.value

    def get_leftChild(self):
        return self.leftChild    

    def get_rightChild(self):
        return self.rightChild

    def set_leftChild(self, key, value):
        node = Node(key, value, None, None, self)
        self.leftChild = node

    def set_rightChild(self, key, value):
        node = Node(key, value, None, None, self)
        self.rightChild = node 

    def isLeaf(self):
        if self.leftChild is None and self.rightChild is None:
           return True
        else:
           return False   

class BinaryTree:
    def __init__(self, key = None, value = None, leftChild = None, rightChild = None, parent = None):
        self.root = Node(key, value, leftChild = None, rightChild = None, parent = None)

    def addChild(self, key, value):
        current = self.root
        if current is None:
           current = Node(key, value, leftChild = None, rightChild = None, parent = None)
        else:   
           self._addChild(current, key, value)

    def _addChild(self, current, key, value):                
        if current is None:
           current = Node(key, value, leftChild = None, rightChild = None, parent = current)
           return 

        if current.isLeaf() is True:
            if current.get_key() > key:
               current.set_leftChild(key, value)
            else:
               current.set_rightChild(key, value)
            return
        elif current.get_key() > key:
                return self._addChild(current.get_leftChild(), key, value) 
        elif current.get_key() <= key:                
                return self._addChild(current.get_rightChild(), key, value)

    def traversalInOrder(self):
        if self.root is None:
            return None 
        else:   
            self._traversalInOrder(self.root)            

obj = BinaryTree(10, "ram")
obj.addChild(12, "hari")
obj.addChild(4, "hans")

Answer 1

当current is None时，您实际上并未将值存储在树中，而只存储在局部变量中。

这样它应该有效：

class BinaryTree:
    def __init__(self):
        self.root = None

    def addChild(self, key, value):
        current = self.root
        if current is None:
          self.root = Node(key, value, leftChild = None, rightChild = None, parent = None)
        else:  
          self._addChild(current, key, value)

    def _addChild(self, current, key, value):                
        if current.get_key() > key:
          child = current.get_leftChild()
          if child is None:
            current.set_leftChild(key, value)
          else:
            self._addChild(child, key, value) 
        else:
          child = current.get_rightChild()
          if child is None:
            current.set_rightChild(key, value)
          else:
            self._addChild(child, key, value) 


obj = BinaryTree()
obj.addChild(10, "ram")
obj.addChild(12, "hari")
obj.addChild(4, "hans")

python中的二叉搜索树implemntation

1 个答案: