In B⁺ tree each key is stored in a leaf node. So, the order of leaf node in B⁺ tree is the maximum number of (key, record pointer)pairs it can hold. What is the order of B⁺ tree leaf node, when block size is given as 4 Kilo byte record pointer(R_P) of 8 byte, block pointer(B_P) size is 2 byte and key is of 15 byte.

(A) 164
(B) 178
(C) 179
(D) 241


Answer: (B)

Explanation: Let order of leaf node of B⁺ tree as P.
P(key + R_P) B_P ≤ block size.
P(15 + 8) + 2 ≤ 4 * 1024
23 * P ≤ 4094
P ≤ 4094 / 23
P ≤ 178
So, option (A) is correct.


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