B+ Tree

by anupmaurya

A B+ tree follows a multi-level index format and is a balanced binary search tree. A B+ tree’s leaf nodes denote actual data pointers. The B+ tree ensures that all leaf nodes stay balanced at the same height. In addition, a link list connects the leaf nodes; a B+ tree can therefore allow both random access and sequential access.

Structure of B+ Tree

B+ Tree
Structure of B+ Tree

Each node on the leaf is the same distance from the root node. For any B+ tree, a B+ tree is of the order n where n is fixed.

Internal nodes −

  • Internal (non-leaf) nodes, except the root node, contain at least of [ n/2 ] pointers
  • At most, n pointers may contain an internal nod

Leaf nodes −

  • At least [ n/2 ] record pointers and [n/2] key values are stored in Leaf nodes.
  • At most, n record pointers and n key values will contain a leaf node.
  • Each node of the leaf contains a block pointer P to point to the next node of the leaf and forms a linked list.

B+ Tree Insertion

  • From the bottom, B+ trees are filled and every entry is done at the leaf node.
  • If a leaf node overflows −
    • Split node into two parts.
    • Partition at i = ⌊(m+1)/2⌋.
    • First i entries are stored in one node.
    • Rest of the entries (i+1 onwards) are moved to a new node.
    • ith key is duplicated at the parent of the leaf.
  • if a non-leaf node overflows −
    • Split node into two parts.
    • Partition the node at i = ⌈(m+1)/2⌉.
    • Entries up to i are kept in one node.
    • Rest of the entries are moved to a new node.

B+ Tree Deletion

  • At the leaf nodes, B+ tree entries are deleted.
  • The entry for the goal is searched and deleted.
    • If it is an internal node, delete it from the left position and replace it with an entry.
  • The underflow is tested after deletion,
    • If there is an underflow, allocate the entries from the nodes left over.
  • If distribution from the left is not possible, then
    • Distributing right to it from the nodes.
  • If distribution from the left or from the right is not feasible, then
    • Merge to the left and right of the node.

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