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In a balanced binary search tree with n elements, what is the worst case time complexity of reporting all elements in range [a, b]? Assume that the number of reported elements is k.
- a)Θ(log n)
- b)Θ(n log k)
- c)Θ(log n + k)
- d)Θ(k log n)
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a balanced binary search tree with n elements, what is the worst ca...
Let a = 16, b = 42
16: Find the ‘16’ element in the BST = Ο(logn)
42: Find the ‘42’ element in the BST = Ο(logn)
[16, 42]: Inorder sorted element between 16 to 42
{16, 18, 19, 20, 25, 30, 35, 40, 42} ⇒ requires Ο(k) time for k element
So total time: Ο(2 logn + k) ≡ Ο(logn + k)
Most Upvoted Answer
In a balanced binary search tree with n elements, what is the worst ca...
The worst case time complexity of reporting all elements in the range [a, b] in a balanced binary search tree with n elements is O(k + log n), where k is the number of reported elements.
In the worst case, we might have to traverse down the tree from the root to the lowest common ancestor of nodes with values a and b, which takes O(log n) time in a balanced binary search tree.
Once we reach the lowest common ancestor, we can perform an in-order traversal to report all elements within the range [a, b]. In the worst case, this traversal will visit k nodes, resulting in a time complexity of O(k).
Therefore, the overall worst case time complexity is O(k + log n).
In the worst case, we might have to traverse down the tree from the root to the lowest common ancestor of nodes with values a and b, which takes O(log n) time in a balanced binary search tree.
Once we reach the lowest common ancestor, we can perform an in-order traversal to report all elements within the range [a, b]. In the worst case, this traversal will visit k nodes, resulting in a time complexity of O(k).
Therefore, the overall worst case time complexity is O(k + log n).
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