Python Data Structure and Algorithm Tutorial – B+ Tree

B+ Tree

 

In this tutorial, you will learn what a B+ tree is. Also, you will find working examples of searching operation on a B+ tree in Python.

A B+ tree is an advanced form of a self-balancing tree in which all the values are present in the leaf level.

An important concept to be understood before learning B+ tree is multilevel indexing. In multilevel indexing, the index of indices is created as in figure below. It makes accessing the data easier and faster.

Multilevel Indexing using B+ tree
Multilevel Indexing using B+ tree

Properties of a B+ Tree

  1. All leaves are at the same level.
  2. The root has at least two children.
  3. Each node except root can have a maximum of m children and at least m/2 children.
  4. Each node can contain a maximum of m – 1 keys and a minimum of ⌈m/2⌉ – 1 keys.

 


Comparison between a B-tree and a B+ Tree

 

B-tree
B-tree
B+ tree
B+ tree

The data pointers are present only at the leaf nodes on a B+ tree whereas the data pointers are present in the internal, leaf or root nodes on a B-tree.

The leaves are not connected with each other on a B-tree whereas they are connected on a B+ tree.

Operations on a B+ tree are faster than on a B-tree.


The following steps are followed to search for data in a B+ Tree of order m. Let the data to be searched be k.

  1. Start from the root node. Compare k with the keys at the root node [k1, k2, k3,……km – 1].
  2. If k < k1, go to the left child of the root node.
  3. Else if k == k1, compare k2. If k < k2, k lies between k1 and k2. So, search in the left child of k2.
  4. If k > k2, go for k3, k4,…km-1 as in steps 2 and 3.
  5. Repeat the above steps until a leaf node is reached.
  6. If k exists in the leaf node, return true else return false.

 


Searching Example on a B+ Tree

Let us search k = 45 on the following B+ tree.

B+ tree
B+ tree
  1. Compare k with the root node.
    B+ tree search
    k is not found at the root
  2. Since k > 25, go to the right child.
    B+ tree search
    Go to right of the root
  3. Compare k with 35. Since k > 30, compare k with 45.
    B+ tree search
    k not found
  4. Since k ≥ 45, so go to the right child.
    B+ tree search
    go to the right
  5. k is found.
    B+ tree search
    k is found

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")



Search Complexity

Time Complexity

If linear search is implemented inside a node, then total complexity is Θ(logt n).

If binary search is used, then total complexity is Θ(log2t.logt n).


B+ Tree Applications

  • Multilevel Indexing
  • Faster operations on the tree (insertion, deletion, search)
  • Database indexing

Python Example for Beginners

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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.

 

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