Last Updated : 28 Dec, 2023
Comments
Improve
The limitations of traditional binary search trees can be frustrating. Meet the B-Tree, the multi-talented data structure that can handle massive amounts of data with ease. When it comes to storing and searching large amounts of data, traditional binary search trees can become impractical due to their poor performance and high memory usage. B-Trees, also known as B-Tree or Balanced Tree, are a type of self-balancing tree that was specifically designed to overcome these limitations.
Unlike traditional binary search trees, B-Trees are characterized by the large number of keys that they can store in a single node, which is why they are also known as “large key” trees. Each node in a B-Tree can contain multiple keys, which allows the tree to have a larger branching factor and thus a shallower height. This shallow height leads to less disk I/O, which results in faster search and insertion operations. B-Trees are particularly well suited for storage systems that have slow, bulky data access such as hard drives, flash memory, and CD-ROMs.
B-Trees maintains balance by ensuring that each node has a minimum number of keys, so the tree is always balanced. This balance guarantees that the time complexity for operations such as insertion, deletion, and searching is always O(log n), regardless of the initial shape of the tree.
Time Complexity of B-Tree:
Sr. No. | Algorithm | Time Complexity |
---|---|---|
1. | Search | O(log n) |
2. | Insert | O(log n) |
3. | Delete | O(log n) |
Note: “n” is the total number of elements in the B-tree
Properties of B-Tree:
- All leaves are at the same level.
- B-Tree is defined by the term minimum degree ‘t‘. The value of ‘t‘ depends upon disk block size.
- Every node except the root must contain at least t-1 keys. The root may contain a minimum of 1 key.
- All nodes (including root) may contain at most (2*t – 1) keys.
- Number of children of a node is equal to the number of keys in it plus 1.
- All keys of a node are sorted in increasing order. The child between two keys k1 and k2 contains all keys in the range from k1 and k2.
- B-Tree grows and shrinks from the root which is unlike Binary Search Tree. Binary Search Trees grow downward and also shrink from downward.
- Like other balanced Binary Search Trees, the time complexity to search, insert, and delete is O(log n).
- Insertion of a Node in B-Tree happens only at Leaf Node.
Following is an example of a B-Tree of minimum order 5
Note: that in practical B-Trees, the value of the minimum order is much more than 5.
We can see in the above diagram that all the leaf nodes are at the same level and all non-leafs have no empty sub-tree and have keys one less than the number of their children.
Interesting Facts about B-Trees:
- The minimum height of the B-Tree that can exist with n number of nodes and m is the maximum number of children of a node can have is:
- The maximum height of the B-Tree that can exist with n number of nodes and t is the minimum number of children that a non-root node can have is:and
Traversal in B-Tree:
Traversal is also similar to Inorder traversal of Binary Tree. We start from the leftmost child, recursively print the leftmost child, then repeat the same process for the remaining children and keys. In the end, recursively print the rightmost child.
Search Operation in B-Tree:
Search is similar to the search in Binary Search Tree. Let the key to be searched is k.
- Start from the root and recursively traverse down.
- For every visited non-leaf node,
- If the node has the key, we simply return the node.
- Otherwise, we recur down to the appropriate child (The child which is just before the first greater key) of the node.
- If we reach a leaf node and don’t find k in the leaf node, then return NULL.
Searching a B-Tree is similar to searching a binary tree. The algorithm is similar and goes with recursion. At each level, the search is optimized as if the key value is not present in the range of the parent then the key is present in another branch. As these values limit the search they are also known as limiting values or separation values. If we reach a leaf node and don’t find the desired key then it will display NULL.
Algorithm for Searching an Element in a B-Tree:-
C++
struct
Node {
int
n;
int
key[MAX_KEYS];
Node* child[MAX_CHILDREN];
bool
leaf;
};
Node* BtreeSearch(Node* x,
int
k) {
int
i = 0;
while
(i < x->n && k > x->key[i]) {
i++;
}
if
(i < x->n && k == x->key[i]) {
return
x;
}
if
(x->leaf) {
return
nullptr;
}
return
BtreeSearch(x->child[i], k);
}
C
BtreeSearch(x, k)
i = 1
// n[x] means number of keys in x node
while
i ? n[x] and k ? keyi[x]
do
i = i + 1
if
i n[x] and k = keyi[x]
then
return
(x, i)
if
leaf [x]
then
return
NIL
else
return
BtreeSearch(ci[x], k)
Java
class
Node {
int
n;
int
[] key =
new
int
[MAX_KEYS];
Node[] child =
new
Node[MAX_CHILDREN];
boolean
leaf;
}
Node BtreeSearch(Node x,
int
k) {
int
i =
0
;
while
(i < x.n && k >= x.key[i]) {
i++;
}
if
(i < x.n && k == x.key[i]) {
return
x;
}
if
(x.leaf) {
return
null
;
}
return
BtreeSearch(x.child[i], k);
}
Python3
class
Node:
def
__init__(
self
):
self
.n
=
0
self
.key
=
[
0
]
*
MAX_KEYS
self
.child
=
[
None
]
*
MAX_CHILDREN
self
.leaf
=
True
def
BtreeSearch(x, k):
i
=
0
while
i < x.n
and
k >
=
x.key[i]:
i
+
=
1
if
i < x.n
and
k
=
=
x.key[i]:
return
x
if
x.leaf:
return
None
return
BtreeSearch(x.child[i], k)
C#
class
Node {
public
int
n;
public
int
[] key =
new
int
[MAX_KEYS];
public
Node[] child =
new
Node[MAX_CHILDREN];
public
bool
leaf;
}
Node BtreeSearch(Node x,
int
k) {
int
i = 0;
while
(i < x.n && k >= x.key[i]) {
i++;
}
if
(i < x.n && k == x.key[i]) {
return
x;
}
if
(x.leaf) {
return
null
;
}
return
BtreeSearch(x.child[i], k);
}
Javascript
// Define a Node class with properties n, key, child, and leaf
class Node {
constructor() {
this
.n = 0;
this
.key =
new
Array(MAX_KEYS);
this
.child =
new
Array(MAX_CHILDREN);
this
.leaf =
false
;
}
}
// Define a function BtreeSearch that takes in a Node object x and an integer k
function
BtreeSearch(x, k) {
let i = 0;
while
(i < x.n && k >= x.key[i]) {
i++;
}
if
(i < x.n && k == x.key[i]) {
return
x;
}
if
(x.leaf) {
return
null
;
}
return
BtreeSearch(x.child[i], k);
}
Examples:
Input: Search 120 in the given B-Tree.
Solution:
In this example, we can see that our search was reduced by just limiting the chances where the key containing the value could be present. Similarly if within the above example we’ve to look for 180, then the control will stop at step 2 because the program will find that the key 180 is present within the current node. And similarly, if it’s to seek out 90 then as 90 < 100 so it’ll go to the left subtree automatically, and therefore the control flow will go similarly as shown within the above example.
Below is the implementation of the above approach:
C++
// C++ implementation of search() and traverse() methods
#include <iostream>
using
namespace
std;
// A BTree node
class
BTreeNode {
int
* keys;
// An array of keys
int
t;
// Minimum degree (defines the range for number
// of keys)
BTreeNode** C;
// An array of child pointers
int
n;
// Current number of keys
bool
leaf;
// Is true when node is leaf. Otherwise false
public
:
BTreeNode(
int
_t,
bool
_leaf);
// Constructor
// A function to traverse all nodes in a subtree rooted
// with this node
void
traverse();
// A function to search a key in the subtree rooted with
// this node.
BTreeNode*
search(
int
k);
// returns NULL if k is not present.
// Make the BTree friend of this so that we can access
// private members of this class in BTree functions
friend
class
BTree;
};
// A BTree
class
BTree {
BTreeNode* root;
// Pointer to root node
int
t;
// Minimum degree
public
:
// Constructor (Initializes tree as empty)
BTree(
int
_t)
{
root = NULL;
t = _t;
}
// function to traverse the tree
void
traverse()
{
if
(root != NULL)
root->traverse();
}
// function to search a key in this tree
BTreeNode* search(
int
k)
{
return
(root == NULL) ? NULL : root->search(k);
}
};
// Constructor for BTreeNode class
BTreeNode::BTreeNode(
int
_t,
bool
_leaf)
{
// Copy the given minimum degree and leaf property
t = _t;
leaf = _leaf;
// Allocate memory for maximum number of possible keys
// and child pointers
keys =
new
int
[2 * t - 1];
C =
new
BTreeNode*[2 * t];
// Initialize the number of keys as 0
n = 0;
}
// Function to traverse all nodes in a subtree rooted with
// this node
void
BTreeNode::traverse()
{
// There are n keys and n+1 children, traverse through n
// keys and first n children
int
i;
for
(i = 0; i < n; i++) {
// If this is not leaf, then before printing key[i],
// traverse the subtree rooted with child C[i].
if
(leaf ==
false
)
C[i]->traverse();
cout <<
" "
<< keys[i];
}
// Print the subtree rooted with last child
if
(leaf ==
false
)
C[i]->traverse();
}
// Function to search key k in subtree rooted with this node
BTreeNode* BTreeNode::search(
int
k)
{
// Find the first key greater than or equal to k
int
i = 0;
while
(i < n && k > keys[i])
i++;
// If the found key is equal to k, return this node
if
(keys[i] == k)
return
this
;
// If the key is not found here and this is a leaf node
if
(leaf ==
true
)
return
NULL;
// Go to the appropriate child
return
C[i]->search(k);
}
Java
// Java program to illustrate the sum of two numbers
// A BTree
class
Btree {
public
BTreeNode root;
// Pointer to root node
public
int
t;
// Minimum degree
// Constructor (Initializes tree as empty)
Btree(
int
t)
{
this
.root =
null
;
this
.t = t;
}
// function to traverse the tree
public
void
traverse()
{
if
(
this
.root !=
null
)
this
.root.traverse();
System.out.println();
}
// function to search a key in this tree
public
BTreeNode search(
int
k)
{
if
(
this
.root ==
null
)
return
null
;
else
return
this
.root.search(k);
}
}
// A BTree node
class
BTreeNode {
int
[] keys;
// An array of keys
int
t;
// Minimum degree (defines the range for number
// of keys)
BTreeNode[] C;
// An array of child pointers
int
n;
// Current number of keys
boolean
leaf;
// Is true when node is leaf. Otherwise false
// Constructor
BTreeNode(
int
t,
boolean
leaf)
{
this
.t = t;
this
.leaf = leaf;
this
.keys =
new
int
[
2
* t -
1
];
this
.C =
new
BTreeNode[
2
* t];
this
.n =
0
;
}
// A function to traverse all nodes in a subtree rooted
// with this node
public
void
traverse()
{
// There are n keys and n+1 children, traverse
// through n keys and first n children
int
i =
0
;
for
(i =
0
; i <
this
.n; i++) {
// If this is not leaf, then before printing
// key[i], traverse the subtree rooted with
// child C[i].
if
(
this
.leaf ==
false
) {
C[i].traverse();
}
System.out.print(keys[i] +
" "
);
}
// Print the subtree rooted with last child
if
(leaf ==
false
)
C[i].traverse();
}
// A function to search a key in the subtree rooted with
// this node.
BTreeNode search(
int
k)
{
// returns NULL if k is not present.
// Find the first key greater than or equal to k
int
i =
0
;
while
(i < n && k > keys[i])
i++;
// If the found key is equal to k, return this node
if
(keys[i] == k)
return
this
;
// If the key is not found here and this is a leaf
// node
if
(leaf ==
true
)
return
null
;
// Go to the appropriate child
return
C[i].search(k);
}
}
Python3
# Create a node
class
BTreeNode:
def
__init__(
self
, leaf
=
False
):
self
.leaf
=
leaf
self
.keys
=
[]
self
.child
=
[]
# Tree
class
BTree:
def
__init__(
self
, t):
self
.root
=
BTreeNode(
True
)
self
.t
=
t
# Insert node
def
insert(
self
, k):
root
=
self
.root
if
len
(root.keys)
=
=
(
2
*
self
.t)
-
1
:
temp
=
BTreeNode()
self
.root
=
temp
temp.child.insert(
0
, root)
self
.split_child(temp,
0
)
self
.insert_non_full(temp, k)
else
:
self
.insert_non_full(root, k)
# Insert nonfull
def
insert_non_full(
self
, x, k):
i
=
len
(x.keys)
-
1
if
x.leaf:
x.keys.append((
None
,
None
))
while
i >
=
0
and
k[
0
] < x.keys[i][
0
]:
x.keys[i
+
1
]
=
x.keys[i]
i
-
=
1
x.keys[i
+
1
]
=
k
else
:
while
i >
=
0
and
k[
0
] < x.keys[i][
0
]:
i
-
=
1
i
+
=
1
if
len
(x.child[i].keys)
=
=
(
2
*
self
.t)
-
1
:
self
.split_child(x, i)
if
k[
0
] > x.keys[i][
0
]:
i
+
=
1
self
.insert_non_full(x.child[i], k)
# Split the child
def
split_child(
self
, x, i):
t
=
self
.t
y
=
x.child[i]
z
=
BTreeNode(y.leaf)
x.child.insert(i
+
1
, z)
x.keys.insert(i, y.keys[t
-
1
])
z.keys
=
y.keys[t: (
2
*
t)
-
1
]
y.keys
=
y.keys[
0
: t
-
1
]
if
not
y.leaf:
z.child
=
y.child[t:
2
*
t]
y.child
=
y.child[
0
: t
-
1
]
# Print the tree
def
print_tree(
self
, x, l
=
0
):
print
(
"Level "
, l,
" "
,
len
(x.keys), end
=
":"
)
for
i
in
x.keys:
print
(i, end
=
" "
)
print
()
l
+
=
1
if
len
(x.child) >
0
:
for
i
in
x.child:
self
.print_tree(i, l)
# Search key in the tree
def
search_key(
self
, k, x
=
None
):
if
x
is
not
None
:
i
=
0
while
i <
len
(x.keys)
and
k > x.keys[i][
0
]:
i
+
=
1
if
i <
len
(x.keys)
and
k
=
=
x.keys[i][
0
]:
return
(x, i)
elif
x.leaf:
return
None
else
:
return
self
.search_key(k, x.child[i])
else
:
return
self
.search_key(k,
self
.root)
def
main():
B
=
BTree(
3
)
for
i
in
range
(
10
):
B.insert((i,
2
*
i))
B.print_tree(B.root)
if
B.search_key(
8
)
is
not
None
:
print
(
"\nFound"
)
else
:
print
(
"\nNot Found"
)
if
__name__
=
=
'__main__'
:
main()
C#
// C# program to illustrate the sum of two numbers
using
System;
// A BTree
class
Btree {
public
BTreeNode root;
// Pointer to root node
public
int
t;
// Minimum degree
// Constructor (Initializes tree as empty)
Btree(
int
t)
{
this
.root =
null
;
this
.t = t;
}
// function to traverse the tree
public
void
traverse()
{
if
(
this
.root !=
null
)
this
.root.traverse();
Console.WriteLine();
}
// function to search a key in this tree
public
BTreeNode search(
int
k)
{
if
(
this
.root ==
null
)
return
null
;
else
return
this
.root.search(k);
}
}
// A BTree node
class
BTreeNode {
int
[] keys;
// An array of keys
int
t;
// Minimum degree (defines the range for number
// of keys)
BTreeNode[] C;
// An array of child pointers
int
n;
// Current number of keys
bool
leaf;
// Is true when node is leaf. Otherwise false
// Constructor
BTreeNode(
int
t,
bool
leaf)
{
this
.t = t;
this
.leaf = leaf;
this
.keys =
new
int
[2 * t - 1];
this
.C =
new
BTreeNode[2 * t];
this
.n = 0;
}
// A function to traverse all nodes in a subtree rooted
// with this node
public
void
traverse()
{
// There are n keys and n+1 children, traverse
// through n keys and first n children
int
i = 0;
for
(i = 0; i <
this
.n; i++) {
// If this is not leaf, then before printing
// key[i], traverse the subtree rooted with
// child C[i].
if
(
this
.leaf ==
false
) {
C[i].traverse();
}
Console.Write(keys[i] +
" "
);
}
// Print the subtree rooted with last child
if
(leaf ==
false
)
C[i].traverse();
}
// A function to search a key in the subtree rooted with
// this node.
public
BTreeNode search(
int
k)
{
// returns NULL if k is not present.
// Find the first key greater than or equal to k
int
i = 0;
while
(i < n && k > keys[i])
i++;
// If the found key is equal to k, return this node
if
(keys[i] == k)
return
this
;
// If the key is not found here and this is a leaf
// node
if
(leaf ==
true
)
return
null
;
// Go to the appropriate child
return
C[i].search(k);
}
}
// This code is contributed by Rajput-Ji
Javascript
// Javascript program to illustrate the sum of two numbers
// A BTree
class Btree
{
// Constructor (Initializes tree as empty)
constructor(t)
{
this
.root =
null
;
this
.t = t;
}
// function to traverse the tree
traverse()
{
if
(
this
.root !=
null
)
this
.root.traverse();
document.write(
"<br>"
);
}
// function to search a key in this tree
search(k)
{
if
(
this
.root ==
null
)
return
null
;
else
return
this
.root.search(k);
}
}
// A BTree node
class BTreeNode
{
// Constructor
constructor(t,leaf)
{
this
.t = t;
this
.leaf = leaf;
this
.keys =
new
Array(2 * t - 1);
this
.C =
new
Array(2 * t);
this
.n = 0;
}
// A function to traverse all nodes in a subtree rooted with this node
traverse()
{
// There are n keys and n+1 children, traverse through n keys
// and first n children
let i = 0;
for
(i = 0; i <
this
.n; i++) {
// If this is not leaf, then before printing key[i],
// traverse the subtree rooted with child C[i].
if
(
this
.leaf ==
false
) {
C[i].traverse();
}
document.write(keys[i] +
" "
);
}
// Print the subtree rooted with last child
if
(leaf ==
false
)
C[i].traverse();
}
// A function to search a key in the subtree rooted with this node.
search(k)
// returns NULL if k is not present.
{
// Find the first key greater than or equal to k
let i = 0;
while
(i < n && k > keys[i])
i++;
// If the found key is equal to k, return this node
if
(keys[i] == k)
return
this
;
// If the key is not found here and this is a leaf node
if
(leaf ==
true
)
return
null
;
// Go to the appropriate child
return
C[i].search(k);
}
}
// This code is contributed by patel2127
Note: The above code doesn’t contain the driver program. We will be covering the complete program in our next post on B-Tree Insertion.
There are two conventions to define a B-Tree, one is to define by minimum degree, second is to define by order. We have followed the minimum degree convention and will be following the same in coming posts on B-Tree. The variable names used in the above program are also kept the same
Applications of B-Trees:
- It is used in large databases to access data stored on the disk
- Searching for data in a data set can be achieved in significantly less time using the B-Tree
- With the indexing feature, multilevel indexing can be achieved.
- Most of the servers also use the B-tree approach.
- B-Trees are used in CAD systems to organize and search geometric data.
- B-Trees are also used in other areas such as natural language processing, computer networks, and cryptography.
Advantages of B-Trees:
- B-Trees have a guaranteed time complexity of O(log n) for basic operations like insertion, deletion, and searching, which makes them suitable for large data sets and real-time applications.
- B-Trees are self-balancing.
- High-concurrency and high-throughput.
- Efficient storage utilization.
Disadvantages of B-Trees:
- B-Trees are based on disk-based data structures and can have a high disk usage.
- Not the best for all cases.
- Slow in comparison to other data structures.
Insertion and Deletion:
B-Tree Insertion
B-Tree Deletion
Elevate your coding journey with a Premium subscription. Benefit from ad-free learning, unlimited article summaries, an AI bot, access to 35+ courses, and more-available only with GeeksforGeeks Premium! Explore now!
${secondsToHms(duration)}
${getTagsString(tags)}
'; } } return videos_content; } //function to show main video content with related videos content async function showMainVideoContent(main_video, course_link){ //Load main video $(".video-main").html(`
`); require(["ima"], function() { var player = videojs('article-video', { controls: true, // autoplay: true, // muted: true, controlBar: { pictureInPictureToggle: false }, playbackRates: [0.5, 0.75, 1, 1.25, 1.5, 2], poster: main_video['meta']['largeThumbnail'], sources: [{src: main_video['source'], type: 'application/x-mpegURL'}], tracks: [{src: main_video['subtitle'], kind:'captions', srclang: 'en', label: 'English', default: true}] },function() { player.qualityLevels(); try { player.hlsQualitySelector(); } catch (error) { console.log("HLS not working - ") } document.getElementById('article-video-tab-content').style.display = 'block'; } ); const video = document.querySelector("video"); const events =[ { 'name':'play', 'callback':()=>{videoPlayCallback(main_video['slug'])} }, ]; events.forEach(event=>{ video.addEventListener(event.name,event.callback); }); // error handling for no compatible source player.on('error', function() { var error = player.error(); console.log("Video Error: ", error); if (error && error.code === 4) { console.log("No compatible source was found for this media."); hideVideoPlayer(); } }); }, function(err) { var player = videojs('article-video'); player.createModal('Something went wrong. Please refresh the page to load the video.'); hideVideoPlayer(); // hiding video in case of timeout (requirejs) console.log(err); }); // function to hide the video player function hideVideoPlayer() { var videoPlayer = document.getElementById('article-video'); if (videoPlayer) { videoPlayer.parentNode.removeChild(videoPlayer); } } /*let video_date = main_video['time']; video_date = video_date.split("/"); video_date = formatDate(video_date[2], video_date[1], video_date[0]); let share_section_content = `
`;*/ let hasLikeBtn = false; // console.log(share_section_content); var data = {}; if(false){ try { if((loginData && loginData.isLoggedIn == true)){ const resp = await fetch(`${API_SCRIPT_URL}logged-in-video-details/${main_video['slug']}/`,{ credentials: 'include' }) if(resp.status == 200 || resp.status == 201){ data = await resp.json(); share_section_content+= `
`; } //Load share section // $(".video-share-section").html(share_section_content); // let exitCond = 0; // const delay = (delayInms) => { // return new Promise(resolve => setTimeout(resolve, delayInms)); // } // while(!loginData){ // let delayres = await delay(1000); // exitCond+=1; // console.log(exitCond); // if(exitCond>5){ // break; // } // } // console.log(loginData); /*if(hasLikeBtn && loginData && loginData.isLoggedIn == true){ setLiked(data.liked) setSaved(data.watchlist) }*/ } catch (error) { console.log(error); } } //Load video content like title, description if(false){ $(".video-content-section").html(`
`:''} `); let related_vidoes = main_video['recommendations']; if(!!videos && videos.length>0){ //Load related videos $(".related-videos-content").html(getvideosContent()); } } //show video content // element = document.getElementById('article-video-tab-content'); // element.style.display = 'block'; $('.spinner-loading-overlay:eq(0)').remove(); $('.spinner-loading-overlay:eq(0)').remove(); } await showMainVideoContent(video_data, course_link); // fitRelatedVideosDescription(); } catch (error) { console.log(error); } } getVideoData(); /* $(window).resize(function(){ onWidthChangeEventsListener(); }); $('#video_nav_tab').click('on', function(){ fitRelatedVideosDescription(); });*/ });