Deletion from a B+ Tree
In this tutorial, you will learn about deletion operation on a B+ tree. Also, you will find working examples of deleting elements from a B+ tree in C, C++, Java and Python.
Deleting an element on a B+ tree consists of three main events: searching the node where the key to be deleted exists, deleting the key and balancing the tree if required.Underflow is a situation when there is less number of keys in a node than the minimum number of keys it should hold.
Deletion Operation
Before going through the steps below, one must know these facts about a B+ tree of degree m.
- A node can have a maximum of m children. (i.e. 3)
- A node can contain a maximum of
m - 1keys. (i.e. 2) - A node should have a minimum of
⌈m/2⌉children. (i.e. 2) - A node (except root node) should contain a minimum of
⌈m/2⌉ - 1keys. (i.e. 1)
While deleting a key, we have to take care of the keys present in the internal nodes (i.e. indexes) as well because the values are redundant in a B+ tree. Search the key to be deleted then follow the following steps.
Case I
The key to be deleted is present only at the leaf node not in the indexes (or internal nodes). There are two cases for it:
- There is more than the minimum number of keys in the node. Simply delete the key.
Deleting 40 from B-tree - There is an exact minimum number of keys in the node. Delete the key and borrow a key from the immediate sibling. Add the median key of the sibling node to the parent.
Deleting 5 from B-tree
Case II
The key to be deleted is present in the internal nodes as well. Then we have to remove them from the internal nodes as well. There are the following cases for this situation.
- If there is more than the minimum number of keys in the node, simply delete the key from the leaf node and delete the key from the internal node as well.
Fill the empty space in the internal node with the inorder successor.
Deleting 45 from B-tree - If there is an exact minimum number of keys in the node, then delete the key and borrow a key from its immediate sibling (through the parent).
Fill the empty space created in the index (internal node) with the borrowed key.
Deleting 35 from B-tree - This case is similar to Case II(1) but here, empty space is generated above the immediate parent node.
After deleting the key, merge the empty space with its sibling.
Fill the empty space in the grandparent node with the inorder successor.
Deleting 25 from B-tree
Case III
In this case, the height of the tree gets shrinked. It is a little complicated.Deleting 55 from the tree below leads to this condition. It can be understood in the illustrations below.
Python Examples
/* B+ tee in python */
import math
/* Node creation */
class Node:
def __init__(self, order):
self.order = order
self.values = []
self.keys = []
self.nextKey = None
self.parent = None
self.check_leaf = False
/* Insert at the leaf */
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 plus tree */
class BplusTree:
def __init__(self, order):
self.root = Node(order)
self.root.check_leaf = True
/* Insert operation */
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 / 2)) - 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)
/* Search operation for different operations */
def search(self, value):
current_node = self.root
while(current_node.check_leaf == False):
temp2 = current_node.values
for i in range(len(temp2)):
if (value == temp2[i]):
current_node = current_node.keys[i + 1]
break
elif (value < temp2[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
/* Find the node */
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
/* Inserting at the parent */
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 / 2)) - 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)
/* Delete a node */
def delete(self, value, key):
node_ = self.search(value)
temp = 0
for i, item in enumerate(node_.values):
if item == value:
temp = 1
if key in node_.keys[i]:
if len(node_.keys[i]) > 1:
node_.keys[i].pop(node_.keys[i].index(key))
elif node_ == self.root:
node_.values.pop(i)
node_.keys.pop(i)
else:
node_.keys[i].pop(node_.keys[i].index(key))
del node_.keys[i]
node_.values.pop(node_.values.index(value))
self.deleteEntry(node_, value, key)
else:
print("Value not in Key")
return
if temp == 0:
print("Value not in Tree")
return
/* Delete an entry */
def deleteEntry(self, node_, value, key):
if not node_.check_leaf:
for i, item in enumerate(node_.keys):
if item == key:
node_.keys.pop(i)
break
for i, item in enumerate(node_.values):
if item == value:
node_.values.pop(i)
break
if self.root == node_ and len(node_.keys) == 1:
self.root = node_.keys[0]
node_.keys[0].parent = None
del node_
return
elif (len(node_.keys) < int(math.ceil(node_.order / 2)) and node_.check_leaf == False) or (len(node_.values) < int(math.ceil((node_.order - 1) / 2)) and node_.check_leaf == True):
is_predecessor = 0
parentNode = node_.parent
PrevNode = -1
NextNode = -1
PrevK = -1
PostK = -1
for i, item in enumerate(parentNode.keys):
if item == node_:
if i > 0:
PrevNode = parentNode.keys[i - 1]
PrevK = parentNode.values[i - 1]
if i < len(parentNode.keys) - 1:
NextNode = parentNode.keys[i + 1]
PostK = parentNode.values[i]
if PrevNode == -1:
ndash = NextNode
value_ = PostK
elif NextNode == -1:
is_predecessor = 1
ndash = PrevNode
value_ = PrevK
else:
if len(node_.values) + len(NextNode.values) < node_.order:
ndash = NextNode
value_ = PostK
else:
is_predecessor = 1
ndash = PrevNode
value_ = PrevK
if len(node_.values) + len(ndash.values) < node_.order:
if is_predecessor == 0:
node_, ndash = ndash, node_
ndash.keys += node_.keys
if not node_.check_leaf:
ndash.values.append(value_)
else:
ndash.nextKey = node_.nextKey
ndash.values += node_.values
if not ndash.check_leaf:
for j in ndash.keys:
j.parent = ndash
self.deleteEntry(node_.parent, value_, node_)
del node_
else:
if is_predecessor == 1:
if not node_.check_leaf:
ndashpm = ndash.keys.pop(-1)
ndashkm_1 = ndash.values.pop(-1)
node_.keys = [ndashpm] + node_.keys
node_.values = [value_] + node_.values
parentNode = node_.parent
for i, item in enumerate(parentNode.values):
if item == value_:
p.values[i] = ndashkm_1
break
else:
ndashpm = ndash.keys.pop(-1)
ndashkm = ndash.values.pop(-1)
node_.keys = [ndashpm] + node_.keys
node_.values = [ndashkm] + node_.values
parentNode = node_.parent
for i, item in enumerate(p.values):
if item == value_:
parentNode.values[i] = ndashkm
break
else:
if not node_.check_leaf:
ndashp0 = ndash.keys.pop(0)
ndashk0 = ndash.values.pop(0)
node_.keys = node_.keys + [ndashp0]
node_.values = node_.values + [value_]
parentNode = node_.parent
for i, item in enumerate(parentNode.values):
if item == value_:
parentNode.values[i] = ndashk0
break
else:
ndashp0 = ndash.keys.pop(0)
ndashk0 = ndash.values.pop(0)
node_.keys = node_.keys + [ndashp0]
node_.values = node_.values + [ndashk0]
parentNode = node_.parent
for i, item in enumerate(parentNode.values):
if item == value_:
parentNode.values[i] = ndash.values[0]
break
if not ndash.check_leaf:
for j in ndash.keys:
j.parent = ndash
if not node_.check_leaf:
for j in node_.keys:
j.parent = node_
if not parentNode.check_leaf:
for j in parentNode.keys:
j.parent = parentNode
/* Print the tree */
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('5', '33')
bplustree.insert('15', '21')
bplustree.insert('25', '31')
bplustree.insert('35', '41')
bplustree.insert('45', '10')
printTree(bplustree)
if(bplustree.find('5', '34')):
print("Found")
else:
print("Not found")Python Example for Beginners
Two Machine Learning Fields
There are two sides to machine learning:
- Practical Machine Learning:This is about querying databases, cleaning data, writing scripts to transform data and gluing algorithm and libraries together and writing custom code to squeeze reliable answers from data to satisfy difficult and ill defined questions. It’s the mess of reality.
- Theoretical Machine Learning: This is about math and abstraction and idealized scenarios and limits and beauty and informing what is possible. It is a whole lot neater and cleaner and removed from the mess of reality.
Data Science Resources: Data Science Recipes and Applied Machine Learning Recipes
Introduction to Applied Machine Learning & Data Science for Beginners, Business Analysts, Students, Researchers and Freelancers with Python & R Codes @ Western Australian Center for Applied Machine Learning & Data Science (WACAMLDS) !!!
Latest end-to-end Learn by Coding Recipes in Project-Based Learning:
Applied Statistics with R for Beginners and Business Professionals
Data Science and Machine Learning Projects in Python: Tabular Data Analytics
Data Science and Machine Learning Projects in R: Tabular Data Analytics
Python Machine Learning & Data Science Recipes: Learn by Coding
R Machine Learning & Data Science Recipes: Learn by Coding
Comparing Different Machine Learning Algorithms in Python for Classification (FREE)
Disclaimer: The information and code presented within this recipe/tutorial is only for educational and coaching purposes for beginners and developers. Anyone can practice and apply the recipe/tutorial presented here, but the reader is taking full responsibility for his/her actions. The author (content curator) of this recipe (code / program) has made every effort to ensure the accuracy of the information was correct at time of publication. The author (content curator) does not assume and hereby disclaims any liability to any party for any loss, damage, or disruption caused by errors or omissions, whether such errors or omissions result from accident, negligence, or any other cause. The information presented here could also be found in public knowledge domains.