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2022-05-10 19:50
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python实现二叉搜索树的put()函数,如果键已经在树中,就替换有效载荷,而不是用同一个键插入新节点

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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(selt,key,val):
if selt.root:
selt._put(key,val,selt.root)
else:
selt.root = TreeNode(key,val)
selt.size = selt.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)
elif key == currentNode.key:#如果键已在树中,就替换有效载荷
currentNode.val = val
elif key > currentNode.key:
if currentNode.hasRightChild():
self._put(key,val,currentNode.rightChild)
else:
currentNode.rightChild = TreeNode(key,val,parent=currentNode)

if name=='main':
tree = BinarySearchTree()
tree.put('a',2)
tree.put('b',14)
tree.put('c',9)
tree.put('d',7)
tree.put('c',26)

print(tree['c'])

问题:报错该如何修改?我写的put()函数是否正确?

2条回答 默认 最新

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