Which statement describes a key property of a balanced binary search tree that ensures efficient search?

Which statement describes a key property of a balanced binary search tree that ensures efficient search?

• A. All nodes in a balanced BST are stored in arrays.
• B. Search time is O(n) in a balanced BST.
• C. The height is kept logarithmic to ensure O(log n) search.
• D. It cannot support dynamic insertions.

Answer: (C)

The essential idea is that in a binary search tree, how fast you can search depends on the tree’s height. If the tree is balanced, its height grows only logarithmically with the number of nodes, so the worst-case search time is O(log n). This happens because each comparison lets you cut the remaining search space roughly in half, moving one level down the tree each step. Maintaining a shallow height through rebalancing during insertions and deletions is what keeps searches fast as the tree grows. Storing nodes in an array isn’t what guarantees this efficiency, since a BST can be implemented with pointers and dynamic nodes; the array layout isn’t a defining feature of balance. Saying search time is O(n) would ignore the balancing that keeps height small. And balanced BSTs are designed to support insertions (with rebalancing), so the idea that they can’t handle dynamic insertions isn’t correct.

The essential idea is that in a binary search tree, how fast you can search depends on the tree’s height. If the tree is balanced, its height grows only logarithmically with the number of nodes, so the worst-case search time is O(log n). This happens because each comparison lets you cut the remaining search space roughly in half, moving one level down the tree each step. Maintaining a shallow height through rebalancing during insertions and deletions is what keeps searches fast as the tree grows.

Storing nodes in an array isn’t what guarantees this efficiency, since a BST can be implemented with pointers and dynamic nodes; the array layout isn’t a defining feature of balance. Saying search time is O(n) would ignore the balancing that keeps height small. And balanced BSTs are designed to support insertions (with rebalancing), so the idea that they can’t handle dynamic insertions isn’t correct.