Python 为了让树行走不起作用
我正在尝试用python实践BST树实现,下面是我的代码Python 为了让树行走不起作用,python,algorithm,data-structures,binary-search-tree,Python,Algorithm,Data Structures,Binary Search Tree,我正在尝试用python实践BST树实现,下面是我的代码 import pdb class Node(): def __init__(self, parent=None, key=None): self.parent = parent if parent != None else None self.left = None self.right = None self.key = key if key != None el
import pdb
class Node():
def __init__(self, parent=None, key=None):
self.parent = parent if parent != None else None
self.left = None
self.right = None
self.key = key if key != None else None
class BST():
def __init__(self):
self.root = Node()
def insertKey (self, key):
#pdb.set_trace()
# transverse till we find empty position
if (self.root.key == None):
self.root.key = key
else:
node = self.root
while (node.left != None and node.right != None):
if node.key < key:
node = node.right
else:
node = node.left
#we have node either left or right is empty
if node.key < key:
node.right = Node (node, key)
else:
node.left = Node (node, key)
def inOrder (self, node):
#pdb.set_trace()
if node != None:
self.inOrder (node.left)
print node.key
self.inOrder (node.right)
def printLeft (self, node):
if node != None:
self.printLeft (node)
print node.key
def debugAll (self):
self.inOrder (self.root)
#self.printLeft (self.root)
def fromArray (self, numbers):
srt = sorted(numbers)
print srt
length = len(srt)
mid = length/2
rootEle = srt[mid]
self.insertKey (rootEle)
for i in range (1, mid+1):
try:
#pdb.set_trace()
self.insertKey (srt[mid-i])
self.insertKey (srt[mid+i])
except IndexError:
pass
bst = BST()
bst.fromArray ([1,2,4,3,6,5,10,8,9])
bst.debugAll ()
我试图在插入键时通过pdb进行调试,键被正确插入,但在遍历树时,某些节点被跳过,因为它们被标记为“非类型”。可能是我遗漏了一些语言细节 首先,下面的代码不正确:
while (node.left != None and node.right != None):
if node.key < key:
node = node.right
else:
node = node.left
while(node.left!=无和node.right!=无):
如果node.key
如果左侧或右侧节点不存在,它将停止下降
编辑:如果像这样修改循环,它会工作。可以更好地优化,但这只是一个开始
class Node():
def __init__(self, parent=None, key=None):
self.parent = parent if parent != None else None
self.left = None
self.right = None
self.key = key if key != None else None
class BST():
def __init__(self):
self.root = Node()
def insertKey (self, key):
#pdb.set_trace()
# transverse till we find empty position
if (self.root.key == None):
self.root.key = key
else:
node = self.root
while 1:
if node.key < key:
if node.right is None:
node.right = Node(node, key)
break
else:
node = node.right
else:
if node.left is None:
node.left = Node(node, key)
break
else:
node = node.left
def inOrder (self, node):
#pdb.set_trace()
if node != None:
self.inOrder (node.left)
print node.key
self.inOrder (node.right)
def printLeft (self, node):
if node != None:
self.printLeft (node)
print node.key
def debugAll (self):
self.inOrder (self.root)
#self.printLeft (self.root)
def fromArray (self, numbers):
srt = sorted(numbers)
print srt
length = len(srt)
mid = length/2
rootEle = srt[mid]
self.insertKey (rootEle)
for i in range (1, mid+1):
try:
#pdb.set_trace()
self.insertKey (srt[mid-i])
self.insertKey (srt[mid+i])
except IndexError:
pass
bst = BST()
bst.fromArray ([1,2,4,3,6,5,10,8,9])
bst.debugAll ()
class节点():
定义初始化(self,parent=None,key=None):
self.parent=如果是父级,则为父级!=没有其他的没有
self.left=无
self.right=无
self.key=key如果是key!=没有其他的没有
类BST():
定义初始化(自):
self.root=Node()
def insertKey(自,钥匙):
#pdb.set_trace()
#直到我们找到空位
如果(self.root.key==无):
self.root.key=key
其他:
node=self.root
而1:
如果node.key
您没有正确地形成树,因为它假设要找到一个节点,其中左节点或右节点都不存在,然后,我们将密钥放在新创建的节点中,相应地放在左节点或右节点中。但是如果您找到一个右节点不存在的节点,但您想在左节点中放置一些内容,发生了什么事?@user1064504——我更新了答案,它起作用了,至少在这种情况下是这样。立即想到两种可能的优化——第一种是不需要单独的树和节点类(每个节点都是树的起点),第二种是如果将左/右存储为具有两个值的单个子数组,则可以设置index=node.key
,然后使用node.child[index]在循环中删掉一半的代码。玩得高兴另外,self.parent=parent if parent!=除此之外,None
确实是惯用的,但如果您要存储None,则完全没有必要。只需使用self.parent=None
。检查是否为None
比检查==None
更符合习惯(而且更便宜)。这之所以有效,是因为没有一个是单态的——只能有一个(无)。
class Node():
def __init__(self, parent=None, key=None):
self.parent = parent if parent != None else None
self.left = None
self.right = None
self.key = key if key != None else None
class BST():
def __init__(self):
self.root = Node()
def insertKey (self, key):
#pdb.set_trace()
# transverse till we find empty position
if (self.root.key == None):
self.root.key = key
else:
node = self.root
while 1:
if node.key < key:
if node.right is None:
node.right = Node(node, key)
break
else:
node = node.right
else:
if node.left is None:
node.left = Node(node, key)
break
else:
node = node.left
def inOrder (self, node):
#pdb.set_trace()
if node != None:
self.inOrder (node.left)
print node.key
self.inOrder (node.right)
def printLeft (self, node):
if node != None:
self.printLeft (node)
print node.key
def debugAll (self):
self.inOrder (self.root)
#self.printLeft (self.root)
def fromArray (self, numbers):
srt = sorted(numbers)
print srt
length = len(srt)
mid = length/2
rootEle = srt[mid]
self.insertKey (rootEle)
for i in range (1, mid+1):
try:
#pdb.set_trace()
self.insertKey (srt[mid-i])
self.insertKey (srt[mid+i])
except IndexError:
pass
bst = BST()
bst.fromArray ([1,2,4,3,6,5,10,8,9])
bst.debugAll ()