The following key values are inserted into a B+ tree in which the order of the…

Answer: 4

Reasoning: Leaf nodes can hold at most 2 keys, so any insertion that makes a leaf contain 3 keys causes a leaf split. Simulate the insertions in sequence:

• Insert 10 → single leaf: [10] (no split)

• Insert 3 → leaf becomes [3, 10] (no split)

• Insert 6 → target leaf would be [3, 6, 10], which exceeds capacity → split into [3, 6] and [10]. (leaf splits = 1)

• Insert 8 → goes into [10] → becomes [8, 10] (no split)

• Insert 4 → goes into [3, 6] → becomes [3, 4, 6], overflows → split into [3, 4] and [6]. (leaf splits = 2)

• Insert 2 → goes into [3, 4] → becomes [2, 3, 4], overflows → split into [2, 3] and [4]. (leaf splits = 3). Note: this insertion may cause an internal node to split, but internal splits are not counted here.

• Insert 1 → goes into [2, 3] → becomes [1, 2, 3], overflows → split into [1, 2] and [3]. (leaf splits = 4)

Therefore the maximum number of times leaf nodes get split for this insertion sequence is 4.