Table of Contents
How can a binary search tree be improved?
1 Answer. For such type of data which is in ascending order, You can convert the given BST into a height Balanced binary tree or Self balancing binary tree. That will improve the search operation on the new BST. Infact, Self Balancing binary trees are used to construct and maintain ordered lists such as priority queues …
What is efficiency of binary search tree operations?
Therefore, searching in binary search tree has worst case complexity of O(n). In general, time complexity is O(h) where h is height of BST.
Which of the following is the correct recurrence relation for the binary search algorithm?
Recurrence relation is T(n) = T(n/2) + 1, where T(n) is the time required for binary search in an array of size n. T(n) = T( n 2k )+1+ ··· + 1 Page 2 Since T(1) = 1, when n = 2k, T(n) = T(1) + k = 1 + log2(n). log2(n) ≤ 1 + log2(n) ≤ 2 log2(n) ,∀ n ≥ 2.
When converting binary tree into extended binary tree all the original nodes in binary tree are?
Discussion Forum
| Que. | When converting binary tree into extended binary tree, all the original nodes in binary tree are |
|---|---|
| b. | external nodes on extended tree |
| c. | vanished on extended tree |
| d. | None of above |
| Answer:internal nodes on extended tree |
How to fix a binary search tree (BST)?
Two of the nodes of a Binary Search Tree (BST) are swapped. Fix (or correct) the BST. Input Tree: 10 / \\ 5 8 / \\ 2 20 In the above tree, nodes 20 and 8 must be swapped to fix the tree.
How to correct two swapped nodes of BST are adjacent?
After processing, if the last node value is null, then two swapped nodes of BST are adjacent . Following is the implementation of the given code. // Two nodes in the BST’s swapped, correct the BST. given data and NULL left and right pointers.
How many nodes does the BST tree have?
Swapping 2 and 3 makes the BST valid. The number of nodes in the tree is in the range [2, 1000].
How to store inorder traversal of input tree in a BST?
So a simple method is to store inorder traversal of the input tree in an auxiliary array. Sort the auxiliary array. Finally, insert the auxiliary array elements back to the BST, keeping the structure of the BST same. The time complexity of this method is O (nLogn) and the auxiliary space needed is O (n).