Register

virtual gate

476 views

1 Answer

Best answer
1 votes
1 votes
Option A is the right answer it follows all the properties of B tree with order 3

B tree with order 3 can have max 2 keys and 3 children hence we can see option d and c are wrong also order of 8 9 5 in option c is wrong.

we have A and B left but looking at keys there is 12 missing in B option so we are left with A which satisfied the left biasing also all leaf nodes are at the same level.
• selected by
User Avatar
Voting & Accepting an answer helps improve the community
← PreviousNext →
← Previous in categoryNext in category →

Related questions

0 votes
0 votes
0 answers
1
Sheshang asked Jan 18, 2017
373 views
Virtual Gate Test Series: Databases - B+ Tree
User Avatar
1 votes
1 votes
1 answer
2
casberg asked Nov 23, 2021
527 views
Ace BITS&BYTES, #Trees #B-Tree
What is the smallest and largest number of entries for 2_3 BTree (B2_3 Tree) of height 8 (i.e 8 levels) ? 255 & 6560 127 & 2186 6561 & 255 255 & 2186
What is the smallest and largest number of entries for 2_3 BTree (B2_3 Tree) of height 8 (i.e 8 levels) ? 255 & 6560127 & 21866561 & 255255 & 2186
User Avatar
1 votes
1 votes
1 answer
3
rishu_darkshadow asked Sep 25, 2017
2,022 views
B-Tree
What is the maximum number of nodes and keys in a B-tree of order 10 of depth 3 (root at depth 0) ?
What is the maximum number of nodes and keys in a B-tree of order 10 of depth 3 (root at depth 0) ?
User Avatar
4 votes
4 votes
2 answers
4
go_editor asked Jul 11, 2016
4,462 views
UGC NET CSE | December 2012 | Part 2 | Question: 34
The maximum number of keys stored in a B-tree of order $m$ and depth $d$ is $m^{d +1}-1$ $\frac{m^{d+1}-1}{m-1}$ $(m-1)(m^{d+1}-1)$ $\frac{m^d-1}{m-1}$
The maximum number of keys stored in a B-tree of order $m$ and depth $d$ is$m^{d +1}-1$$\frac{m^{d+1}-1}{m-1}$$(m-1)(m^{d+1}-1)$$\frac{m^d-1}{m-1}$
User Avatar
Link Copied!