Prepare B+ Tree for the following set of key values (2,3,5,7,11,17,19,23,29,31)Assume that the tree is initially empty and values are added in ascending order. Construct B+ tree for the cases where the number of pointers that fit one node is four. After creation of B+ tree perform following series of operations: a)Insert 9 b)Insert 10 c)Insert 8 d)Delete 23 e)Delete 19 ?

Question

Answer

Creating a B-Tree

Assuming an empty tree, start by inserting the first key value, 2, as the root node.

Insert the next key value, 3, to the right of 2.

Insert 5 as the right child of 3.

Since there are already two children of 3, split 3 into a new node and move the middle key, 3, up to the parent node.

Insert 7 as the right child of the newly created node with 3 as the parent.

Insert 11 to the right of 7.

Since there are already two children of the node with 3 as the parent, split it and move the middle key, 7, up to the parent node.

Insert 17 as the right child of the newly created node with 7 as the parent.

Insert 19 to the right of 17.

Since there are already two children of the node with 7 as the parent, split it and move the middle key, 11, up to the parent node.

Insert 23 as the right child of the newly created node with 11 as the parent.

Insert 29 to the right of 23.

Since there are already two children of the node with 11 as the parent, split it and move the middle key, 17, up to the parent node.

Insert 31 as the right child of the newly created node with 17 as the parent.

Performing Operations on the B-Tree

Inserting 9

Start at the root node, 17.

Since 9 is less than the smallest key, 17, follow the left child pointer to the node with 3 as the parent.

Since there is room in the node with 3 as the parent, insert 9 as the right child of 7.

Inserting 10

Start at the root node, 17.

Since 10 is greater than the smallest key, 17, follow the right child pointer to the node with 11 as the parent.

Since there is room in the node with 11 as the parent, insert 10 as the right child of 11.

Inserting 8

Start at the root node, 17.

Since 8 is less than the smallest key, 17, follow the left child pointer to the node with 3 as the parent.

Since there is no room in the node with 3 as the parent, split

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