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Which one of the following is the tightest upper bound that represents the time complexity of
inserting an object into a binary search tree of n nodes?
  • a)
    O(1)
  • b)
    O(log n)
  • c)
    O(n)
  • d)
    O(n log n)
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Which one of the following is the tightest upper bound that represents...
For skewed binary search tree on n nodes, the tightest upper bound to insert a node is O(n)
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Which one of the following is the tightest upper bound that represents...
Time Complexity of Inserting an Object into a Binary Search Tree

To determine the time complexity of inserting an object into a binary search tree of n nodes, we need to analyze the worst-case scenario.

Binary Search Tree Overview
A binary search tree is a data structure in which each node has at most two children: a left child and a right child. The left child is smaller than the parent node, and the right child is greater than the parent node. This property allows for efficient searching, insertion, and deletion operations.

Worst-Case Scenario
The worst-case scenario for inserting an object into a binary search tree occurs when the tree is highly unbalanced and takes the shape of a linked list. In this case, each node only has a right child or a left child, and the tree becomes linear.

Time Complexity Analysis
In the worst-case scenario where the binary search tree is a linked list, the time complexity of inserting an object into the tree would be O(n), where n is the number of nodes in the tree.

Tightest Upper Bound
Among the given options, the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes is O(n).

Explanation
Option a) O(1) represents constant time complexity, which is not possible for inserting an object into a binary search tree since we need to traverse the tree to find the correct position for insertion.

Option b) O(log n) represents logarithmic time complexity and would be valid if the binary search tree is balanced. However, in the worst-case scenario of an unbalanced tree, the time complexity becomes linear.

Option c) O(n) is the correct answer as it represents the tightest upper bound for the worst-case scenario of inserting an object into a binary search tree.

Option d) O(n log n) is a higher time complexity and is not required for the worst-case scenario of inserting an object into a binary search tree.

Therefore, option c) O(n) is the tightest upper bound that represents the time complexity of inserting an object into a binary search tree of n nodes.
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Which one of the following is the tightest upper bound that represents the time complexity ofinserting an object into a binary search tree of n nodes?a)O(1)b)O(log n)c)O(n)d)O(n log n)Correct answer is option 'C'. Can you explain this answer?
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