The time complexity of traversing a binary search tree that contains 'n' elements is O (n) in the worst case. The time complexity of traversing a red-black tree that contains 'n' elements is O (log2 n) in the worst case. Traversing a binary search tree, which contains integers, according to the "inorder" principle always gives us integers in sorted ascending order. If we insert a sequence of integers in an empty BST, it will always be at least as high as the tree we get if we insert the same sequence of integers in an empty red-black tree. If we insert a sequence of integers in an empty BST, it will always be twice as high as the tree we get if we insert the same sequence of integers in an empty red-black tree. Group of answer options Only statements A, B, C and D are correct. Only statements A, B and C are correct. All statements are correct. Only statements A, C, D and E are correct. Only statements A, C and D are correct.
The time complexity of traversing a binary search tree that contains 'n' elements is O (n) in the worst case.
The time complexity of traversing a red-black tree that contains 'n' elements is O (log2 n) in the worst case.
Traversing a binary search tree, which contains integers, according to the "inorder" principle always gives us integers in sorted ascending order.
If we insert a sequence of integers in an empty BST, it will always be at least as high as the tree we get if we insert the same sequence of integers in an empty red-black tree.
If we insert a sequence of integers in an empty BST, it will always be twice as high as the tree we get if we insert the same sequence of integers in an empty red-black tree.
Group of answer options
Only statements A, B, C and D are correct.
Only statements A, B and C are correct.
All statements are correct.
Only statements A, C, D and E are correct.
Only statements A, C and D are correct.
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