
B+树拔出操纵必要思索节点战役衡,怎么是空树,按递删挨次将key拔出叶子节点;怎么没有是空树,须要鉴别索引节点以及叶子节点,没有餍足前提时借要对于节点入止剖析。
Python完成B+树拔出操纵
import math
# 建立节点
class Node:
def __init__(self, order):
self.order = order
self.values = []
self.keys = []
self.nextKey = None
self.parent = None
self.check_leaf = False
def insert_at_leaf(self, leaf, value, key):
if (self.values):
temp1 = self.values
for i in range(len(temp1)):
if (value == temp1[i]):
self.keys[i].append(key)
break
elif (value < temp1[i]):
self.values = self.values[:i] + [value] + self.values[i:]
self.keys = self.keys[:i] + [[key]] + self.keys[i:]
break
elif (i + 1 == len(temp1)):
self.values.append(value)
self.keys.append([key])
break
else:
self.values = [value]
self.keys = [[key]]
# B+树
class BplusTree:
def __init__(self, order):
self.root = Node(order)
self.root.check_leaf = True
# 拔出把持
def insert(self, value, key):
value = str(value)
old_node = self.search(value)
old_node.insert_at_leaf(old_node, value, key)
if (len(old_node.values) == old_node.order):
node1 = Node(old_node.order)
node1.check_leaf = True
node1.parent = old_node.parent
mid = int(math.ceil(old_node.order / 两)) - 1
node1.values = old_node.values[mid + 1:]
node1.keys = old_node.keys[mid + 1:]
node1.nextKey = old_node.nextKey
old_node.values = old_node.values[:mid + 1]
old_node.keys = old_node.keys[:mid + 1]
old_node.nextKey = node1
self.insert_in_parent(old_node, node1.values[0], node1)
# 搜刮操纵
def search(self, value):
current_node = self.root
while(current_node.check_leaf == False):
temp二 = current_node.values
for i in range(len(temp两)):
if (value == temp两[i]):
current_node = current_node.keys[i + 1]
break
elif (value < temp两[i]):
current_node = current_node.keys[i]
break
elif (i + 1 == len(current_node.values)):
current_node = current_node.keys[i + 1]
break
return current_node
# 搜刮节点
def find(self, value, key):
l = self.search(value)
for i, item in enumerate(l.values):
if item == value:
if key in l.keys[i]:
return True
else:
return False
return False
# 正在女级拔出
def insert_in_parent(self, n, value, ndash):
if (self.root == n):
rootNode = Node(n.order)
rootNode.values = [value]
rootNode.keys = [n, ndash]
self.root = rootNode
n.parent = rootNode
ndash.parent = rootNode
return
parentNode = n.parent
temp3 = parentNode.keys
for i in range(len(temp3)):
if (temp3[i] == n):
parentNode.values = parentNode.values[:i] + \
[value] + parentNode.values[i:]
parentNode.keys = parentNode.keys[:i +
1] + [ndash] + parentNode.keys[i + 1:]
if (len(parentNode.keys) > parentNode.order):
parentdash = Node(parentNode.order)
parentdash.parent = parentNode.parent
mid = int(math.ceil(parentNode.order / 二)) - 1
parentdash.values = parentNode.values[mid + 1:]
parentdash.keys = parentNode.keys[mid + 1:]
value_ = parentNode.values[mid]
if (mid == 0):
parentNode.values = parentNode.values[:mid + 1]
else:
parentNode.values = parentNode.values[:mid]
parentNode.keys = parentNode.keys[:mid + 1]
for j in parentNode.keys:
j.parent = parentNode
for j in parentdash.keys:
j.parent = parentdash
self.insert_in_parent(parentNode, value_, parentdash)
# 输入树
def printTree(tree):
lst = [tree.root]
level = [0]
leaf = None
flag = 0
lev_leaf = 0
node1 = Node(str(level[0]) + str(tree.root.values))
while (len(lst) != 0):
x = lst.pop(0)
lev = level.pop(0)
if (x.check_leaf == False):
for i, item in enumerate(x.keys):
print(item.values)
else:
for i, item in enumerate(x.keys):
print(item.values)
if (flag == 0):
lev_leaf = lev
leaf = x
flag = 1
record_len = 3
bplustree = BplusTree(record_len)
bplustree.insert(&#x二7;5&#x两7;, &#x二7;33&#x两7;)
bplustree.insert(&#x二7;15&#x两7;, &#x两7;二1&#x两7;)
bplustree.insert(&#x两7;二5&#x二7;, &#x二7;31&#x两7;)
bplustree.insert(&#x两7;35&#x两7;, &#x两7;41&#x两7;)
bplustree.insert(&#x两7;45&#x二7;, &#x两7;10&#x两7;)
printTree(bplustree)
if(bplustree.find(&#x两7;5&#x两7;, &#x二7;34&#x两7;)):
print("Found")
else:
print("Not found")登录后复造
以上便是用Python编写B+树的拔出操纵的具体形式,更多请存眷萤水红IT仄台此外相闭文章!

发表评论 取消回复