B+ Trees are considered BALANCED because

Data Structures - Quiz(MCQ)

the lengths of the paths from the root to all leaf nodes are all equal.

the number of records in any two leaf nodes differ by at most 1.

the number of children of any two non-leaf sibling nodes differ by at most 1.

the lengths of the paths from the root to all leaf nodes differ from each other by at most 1.

Correct Answer : the lengths of the paths from the root to all leaf nodes are all equal.

Explanation : In both B Tree and B+ trees, depth (length of root to leaf paths) of all leaf nodes is same. This is made sure by the insertion and deletion operations. In these trees, we do insertions in a way that if we have increase height of tree after insertion, we increase height from root. This is different from BST where height increases from leaf nodes. Similarly, if we have to decrease height after deletion, we move the root one level down. This is also different from BST which shrinks from bottom. The above ways of insertion and deletion make sure that depth of every leaf node is same.

Recently Updated in Data Structures Questions

The five items: A, B, C, D, and E are pushed in a stack, one after other starting from A. The stack is popped four items and each element is inserted in a queue. The two elements are deleted from the queue and pushed back on the stack. Now one item is popped from the stack. The popped item is

Correct Answer : D

A three dimensional array in â€˜Câ€™ is declared as int A[x][y][z]. Consider that array elements are stored in row major order and indexing begins from 0. Here, the address of an item at the location A[p][q][r] can be computed as follows (where w is the word length of an integer):

&A[0][0][0] + w(x * y * p + z * q+ r)

&A[0][0][0] + w(y * z * q + z * p + r)

&A[0][0][0] + w(y * z * p + z*q + r)

&A[0][0][0] + w(x * y * q + z * p + r)

Correct Answer : &A[0][0][0] + w(y * z * p + z*q + r)

Explanation : According to above question we have to find the address of A[p][q][r] To reach pth row we must have to cross 0 to p-1 row i.e. p rows and each rows contains y∗z elements Hence , = y∗z∗p Now to reach qth element in pth row we have to cross q rows and each row contains z(total columns) elements =z∗q to reach rth elements we have to cross r elements in (p+1)th row Total elements to cross =(y∗z∗p+z∗q+r) Now each element occupies m amount of space in memory Therefore total space occupied by these elements = m(y∗z∗p+z∗q+r) Hence , address of A[p][q][r]=base address+ Space Occupied by the Elements Before it. =&A[0][0][0]+m(y*z*p+z*q+r) Hence Option (C) is correct.

A B-Tree used as an index for a large database table has four levels including the root node. If a new key is inserted in this index, then the maximum number of nodes that could be newly created in the process are