Final answer:
The average case time complexity for deleting a key from a non-balanced Binary Search Tree with N keys is O(log N). This process includes finding, removing, and potentially rebalancing the tree, which, on average, maintains a log-like structure for such operations.
Step-by-step explanation:
The student's question pertains to the average case time complexity of deleting a key from a non-balanced Binary Search Tree (BST) with N keys. In the average case, this operation has a time complexity of O(log N), assuming that the tree is reasonably balanced, even though it's not guaranteed to be perfectly balanced like in AVL or Red-Black trees. If the tree becomes skewed, the worst case time complexity can degrade to O(N), but on average, especially with random insertion and deletion, the tree maintains a log-like structure.
For deleting a key in a BST, the process typically involves three main steps:
- Finding the element to delete, which takes O(log N) on average if the tree is balanced.
- Removing the element, where you may need to reorganize the tree slightly. If the node is a leaf node or has only one child, it's a simple deletion. However, if the node has two children, you generally find the in-order successor or predecessor and move it to the place of the node being deleted.
- Rebalancing the tree is generally not required for a simple binary search tree upon deletion unless it is a self-balancing BST like AVL or Red-Black tree.
Thus, the correct answer is B: O(log N).