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Insertion in a B+ tree

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  • Difficulty Level : Hard
  • Last Updated : 06 Dec, 2022
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Prerequisite: Introduction of B+ trees
In this article, we will discuss that how to insert a node in B+ Tree. During insertion following properties of B+ Tree must be followed: 

  • Each node except root can have a maximum of M children and at least ceil(M/2) children.
  • Each node can contain a maximum of M – 1 keys and a minimum of ceil(M/2) – 1 keys.
  • The root has at least two children and atleast one search key.
  • While insertion overflow of the node occurs when it contains more than M – 1 search key values.

Here M is the order of B+ tree.
 

Steps for insertion in B+ Tree

  1. Every element is inserted into a leaf node. So, go to the appropriate leaf node.
  2. Insert the key into the leaf node in increasing order only if there is no overflow. If there is an overflow go ahead with the following steps mentioned below to deal with overflow while maintaining the B+ Tree properties.

Properties for insertion B+ Tree

Case 1: Overflow in leaf node 

  1. Split the leaf node into two nodes.
  2. First node contains ceil((m-1)/2) values.
  3. Second node contains the remaining values.
  4. Copy the smallest search key value from second node to the parent node.(Right biased)

Below is the illustration of inserting 8 into B+ Tree of order of 5: 

Case 2: Overflow in non-leaf node 

  1. Split the non leaf node into two nodes.
  2. First node contains ceil(m/2)-1 values.
  3. Move the smallest among remaining to the parent.
  4. Second node contains the remaining keys.

Below is the illustration of inserting 15 into B+ Tree of order of 5: 
 

 

Example to illustrate insertion on a B+ tree

Problem: Insert the following key values 6, 16, 26, 36, 46 on a B+ tree with order = 3. 
Solution: 
Step 1: The order is 3 so at maximum in a node so there can be only 2 search key values. As insertion happens on a leaf node only in a B+ tree so insert search key value 6 and 16 in increasing order in the node. Below is the illustration of the same: 
 

Step 2: We cannot insert 26 in the same node as it causes an overflow in the leaf node, We have to split the leaf node according to the rules. First part contains ceil((3-1)/2) values i.e., only 6. The second node contains the remaining values i.e., 16 and 26. Then also copy the smallest search key value from the second node to the parent node i.e., 16 to the parent node. Below is the illustration of the same: 
 

Step 3: Now the next value is 36 that is to be inserted after 26 but in that node, it causes an overflow again in that leaf node. Again follow the above steps to split the node. First part contains ceil((3-1)/2) values i.e., only 16. The second node contains the remaining values i.e., 26 and 36. Then also copy the smallest search key value from the second node to the parent node i.e., 26 to the parent node. Below is the illustration of the same: 
The illustration is shown in the diagram below. 

Step 4: Now we have to insert 46 which is to be inserted after 36 but it causes an overflow in the leaf node. So we split the node according to the rules. The first part contains 26 and the second part contains 36 and 46 but now we also have to copy 36 to the parent node but it causes overflow as only two search key values can be accommodated in a node. Now follow the steps to deal with overflow in the non-leaf node. 
First node contains ceil(3/2)-1 values i.e. ’16’. 
Move the smallest among remaining to the parent i.e ’26’ will be the new parent node. 
The second node contains the remaining keys i.e ’36’ and the rest of the leaf nodes remain the same. Below is the illustration of the same: 

Below is the python implementation of B+ tree:

Python3




# Python3 program for implementing B+ Tree
 
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)
 
# 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")


Output

['15', '25']
['35', '45']
['5']
Not found

Time complexity: O(log n)
Auxiliary Space: O(log n)

Below is the C++ implementation B+ tree:

C++




#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define pb push_back
 
int bucketSize = 3;
 
// Create 2 classes, one for node and one for tree;
 
class node {
public:
    bool isLeaf;
    node** ptr;
    int *key, size;
    node();
};
node::node()
{
    key = new int[bucketSize];
    ptr = new node*[bucketSize + 1];
}
class Btree {
public:
    // Root of tree stored here;
    node* root;
    Btree();
    void deleteNode(int);
 
    int search(int);
    void display(node*);
    void insert(int);
    node* findParent(node*, node*);
    node* getRoot();
    void shiftLevel(int, node*, node*);
};
 
node* Btree::getRoot() { return root; }
Btree::Btree() { root = NULL; }
 
void Btree::insert(int x)
{
    if (root == NULL) {
        root = new node;
        root->key[0] = x;
        root->isLeaf = true;
        root->size = 1;
    }
 
    else {
        node* current = root;
        node* parent;
 
        while (current->isLeaf == false) {
            parent = current;
 
            for (int i = 0; i < current->size; i++) {
                if (x < current->key[i]) {
                    current = current->ptr[i];
                    break;
                }
 
                if (i == current->size - 1) {
                    current = current->ptr[i + 1];
                    break;
                }
            }
        }
 
        // now we have reached leaf;
        if (current->size
            < bucketSize) { // if the node to be inserted is
                            // not filled
            int i = 0;
 
            // Traverse btree
            while (x > current->key[i] && i < current->size)
                // goto pt where needs to be inserted.
                i++;
 
            for (int j = current->size; j > i; j--)
                // adjust and insert element;
                current->key[j] = current->key[j - 1];
 
            current->key[i] = x;
 
            // size should be increased by 1
            current->size++;
 
            current->ptr[current->size]
                = current->ptr[current->size - 1];
            current->ptr[current->size - 1] = NULL;
        }
 
        // if block does not have enough space;
        else {
            node* newLeaf = new node;
            int tempNode[bucketSize + 1];
 
            for (int i = 0; i < bucketSize; i++)
                // all elements of this block stored
                tempNode[i] = current->key[i];
            int i = 0, j;
 
            // find the right posn of num to be inserted
            while (x > tempNode[i] && i < bucketSize)
                i++;
 
            for (int j = bucketSize + 1; j > i; j--)
                tempNode[j] = tempNode[j - 1];
            tempNode[i] = x;
            // inserted element in its rightful position;
 
            newLeaf->isLeaf = true;
            current->size = (bucketSize + 1) / 2;
            newLeaf->size
                = (bucketSize + 1)
                  - (bucketSize + 1)
                        / 2; // now rearrangement begins!
 
            current->ptr[current->size] = newLeaf;
            newLeaf->ptr[newLeaf->size]
                = current->ptr[bucketSize];
 
            current->ptr[newLeaf->size]
                = current->ptr[bucketSize];
            current->ptr[bucketSize] = NULL;
 
            for (int i = 0; i < current->size; i++)
                current->key[i] = tempNode[i];
 
            for (int i = 0, j = current->size;
                 i < newLeaf->size; i++, j++)
                newLeaf->key[i] = tempNode[j];
 
            // if this is root, then fine,
            // else we need to increase the height of tree;
            if (current == root) {
                node* newRoot = new node;
                newRoot->key[0] = newLeaf->key[0];
                newRoot->ptr[0] = current;
                newRoot->ptr[1] = newLeaf;
                newRoot->isLeaf = false;
                newRoot->size = 1;
                root = newRoot;
            }
            else
                shiftLevel(
                    newLeaf->key[0], parent,
                    newLeaf); // parent->original root
        }
    }
}
 
void Btree::shiftLevel(int x, node* current, node* child)
{ // insert or create an internal node;
    if (current->size
        < bucketSize) { // if can fit in this level, do that
        int i = 0;
        while (x > current->key[i] && i < current->size)
            i++;
        for (int j = current->size; j > i; j--)
            current->key[j] = current->key[j - 1];
 
        for (int j = current->size + 1; j > i + 1; j--)
            current->ptr[j] = current->ptr[j - 1];
 
        current->key[i] = x;
        current->size++;
        current->ptr[i + 1] = child;
    }
 
    // shift up
    else {
        node* newInternal = new node;
        int tempKey[bucketSize + 1];
        node* tempPtr[bucketSize + 2];
 
        for (int i = 0; i < bucketSize; i++)
            tempKey[i] = current->key[i];
 
        for (int i = 0; i < bucketSize + 1; i++)
            tempPtr[i] = current->ptr[i];
 
        int i = 0, j;
        while (x > tempKey[i] && i < bucketSize)
            i++;
 
        for (int j = bucketSize + 1; j > i; j--)
            tempKey[j] = tempKey[j - 1];
 
        tempKey[i] = x;
        for (int j = bucketSize + 2; j > i + 1; j--)
            tempPtr[j] = tempPtr[j - 1];
 
        tempPtr[i + 1] = child;
        newInternal->isLeaf = false;
        current->size = (bucketSize + 1) / 2;
 
        newInternal->size
            = bucketSize - (bucketSize + 1) / 2;
 
        for (int i = 0, j = current->size + 1;
             i < newInternal->size; i++, j++)
            newInternal->key[i] = tempKey[j];
 
        for (int i = 0, j = current->size + 1;
             i < newInternal->size + 1; i++, j++)
            newInternal->ptr[i] = tempPtr[j];
 
        if (current == root) {
            node* newRoot = new node;
            newRoot->key[0] = current->key[current->size];
            newRoot->ptr[0] = current;
            newRoot->ptr[1] = newInternal;
            newRoot->isLeaf = false;
            newRoot->size = 1;
            root = newRoot;
        }
 
        else
            shiftLevel(current->key[current->size],
                       findParent(root, current),
                       newInternal);
    }
}
int Btree::search(int x)
{
    if (root == NULL)
        return -1;
 
    else {
        node* current = root;
        while (current->isLeaf == false) {
            for (int i = 0; i < current->size; i++) {
                if (x < current->key[i]) {
                    current = current->ptr[i];
                    break;
                }
 
                if (i == current->size - 1) {
                    current = current->ptr[i + 1];
                    break;
                }
            }
        }
 
        for (int i = 0; i < current->size; i++) {
            if (current->key[i] == x) {
                // cout<<"Key found "<<endl;
                return 1;
                // return;
            }
        }
 
        // cout<<"Key not found"<<endl;
        return 0;
    }
}
 
// Print the tree
void Btree::display(node* current)
{
    if (current == NULL)
        return;
    queue<node*> q;
    q.push(current);
    while (!q.empty()) {
        int l;
        l = q.size();
 
        for (int i = 0; i < l; i++) {
            node* tNode = q.front();
            q.pop();
 
            for (int j = 0; j < tNode->size; j++)
                if (tNode != NULL)
                    cout << tNode->key[j] << " ";
 
            for (int j = 0; j < tNode->size + 1; j++)
                if (tNode->ptr[j] != NULL)
                    q.push(tNode->ptr[j]);
 
            cout << "\t";
        }
        cout << endl;
    }
}
 
node* Btree::findParent(node* current, node* child)
{
    node* parent;
    if (current->isLeaf || (current->ptr[0])->isLeaf)
        return NULL;
 
    for (int i = 0; i < current->size + 1; i++) {
        if (current->ptr[i] == child) {
            parent = current;
            return parent;
        }
        else {
            parent = findParent(current->ptr[i], child);
            if (parent != NULL)
                return parent;
        }
    }
    return parent;
}
 
signed main()
{
    ios_base::sync_with_stdio(false);
    Btree node;
    cout << "The size of bucket is " << bucketSize << "! "
         << endl;
 
    node.insert(1);
    node.insert(2);
    node.insert(3);
    node.display(node.getRoot());
    node.insert(4);
    node.insert(5);
    node.display(node.getRoot());
 
    return 0;
}


Output

The size of bucket is 3! 
1 2 3     
3     
1 2     3 4 5     
        

Time Complexity:

  • Insertion: O(log (h*bucketSize)), where h is height of the tree, and bucketSize denotes the number of elements that can be stored in a single bucket.
  • Deletion: O(log (h*bucketSize))

Auxiliary Space: O(n), n-> number of elements in the tree.


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