python - 实现二进制搜索树以处理 Python 中的重复键

Question

下面是我发现可以帮助我更好地学习 python 的网站上给出的示例代码:Interactive Python

作者解释说:

One important problem with our implementation of insert is that duplicate keys are not handled properly. As our tree is implemented a duplicate key will create a new node with the same key value in the right subtree of the node having the original key. The result of this is that the node with the new key will never be found during a search. A better way to handle the insertion of a duplicate key is for the value associated with the new key to replace the old value. We leave fixing this bug as an exercise for you."

我的问题是，如何解决此问题以正确处理重复键？如果树中已经存在键，则新的有效负载应替换旧值。目标是不添加具有相同 key 的另一个节点，但我什至不知道从哪里开始这样做。我不确定为什么这会如此令人困惑。

class BinarySearchTree:

    def __init__(self):
        self.root = None
        self.size = 0

    def length(self):
        return self.size

    def __len__(self):
        return self.size

    def __iter__(self):
        return self.root.__iter__()

class TreeNode:
    def __init__(self,key,val,left=None,right=None,
                                       parent=None):
        self.key = key
        self.payload = val
        self.leftChild = left
        self.rightChild = right
        self.parent = parent

    def hasLeftChild(self):
        return self.leftChild

    def hasRightChild(self):
        return self.rightChild

    def isLeftChild(self):
        return self.parent and \
               self.parent.leftChild == self

    def isRightChild(self):
        return self.parent and \
               self.parent.rightChild == self

    def isRoot(self):
        return not self.parent

    def isLeaf(self):
        return not (self.rightChild or self.leftChild)

    def hasAnyChildren(self):
        return self.rightChild or self.leftChild

    def hasBothChildren(self):
        return self.rightChild and self.leftChild

def replaceNodeData(self,key,value,lc,rc):
        self.key = key
        self.payload = value
        self.leftChild = lc
        self.rightChild = rc
        if self.hasLeftChild():
            self.leftChild.parent = self
        if self.hasRightChild():
            self.rightChild.parent = self

    def put(self,key,val):
        if self.root:
            self._put(key,val,self.root)
        else:
            self.root = TreeNode(key,val)
        self.size = self.size + 1

    def _put(self,key,val,currentNode):
        if key < currentNode.key:
            if currentNode.hasLeftChild():
                self._put(key,val,currentNode.leftChild)
            else:
                currentNode.leftChild = TreeNode(key,val,
                                          parent=currentNode)
        else:
            if currentNode.hasRightChild():
                self._put(key,val,currentNode.rightChild)
            else:
                currentNode.rightChild = TreeNode(key,val,
                                          parent=currentNode)

Answer 1

_put() 方法进行测试和插入，它从不测试是否相等。它只测试插入的 key 是否较小(将 key 插入左 child )，否则它会插入右 child 。

简单地测试是否相等并替换:

def _put(self,key,val,currentNode):
    if key == currentNode.key:
        currentNode.value = val

    elif key < currentNode.key:
        if currentNode.hasLeftChild():
            self._put(key,val,currentNode.leftChild)
        else:
            currentNode.leftChild = TreeNode(key,val,
                                      parent=currentNode)
    else:
        if currentNode.hasRightChild():
            self._put(key,val,currentNode.rightChild)
        else:
            currentNode.rightChild = TreeNode(key,val,
                                      parent=currentNode)