4. Given an AVL tree T storing n keys, what are the asymptotic running times, using O notation, for each dictionary operation (i.e., search, insertion, and deletion) on T in the worst-case?
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- Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST. According to the definition of LCA: "The lowest common ancestor is defined between two nodes p and q as the lowest node in t that has both p and q as descendants (where we allow a node to be a descendant of itself)." For example, in the figure from question 1, the LCA between nodes 5 and 46 is 21. You may use the following typedef structure. The function returns the reference of the node that is considered the LCA. typedef struct node_s{ int data; struct node_s * leftchild; struct node _s * rightchild; }node_t;If the linked list pointers were stored in an AVL tree, what would happen to insertions, deletions, and hash table searches? You must explain the running time in terms of the size of the hash table in your answer. Do we need Big-Ohnotation?Given a world map in the form of a Generic M-ary Tree consisting of N nodes and an array of queries[], the task is to implement the functions Lock, Unlock and Upgrade for the given tree. For each query in queries[], the functions return true when the operation is performed successfully, otherwise, it returns false. The functions are defined as: X: Name of the node in the tree and will be uniqueuid: User Id for the person who accesses node X 1. Lock(X, uid): Lock takes exclusive access to the subtree rooted.- Once Lock(X, uid) succeeds, then lock(A, any user) should fail, where A is a descendant of X.- Lock(B. any user) should fail where X is a descendant of B.- Lock operation cannot be performed on a node that is already locked. 2. Unlock(X, uid): To unlock the locked node.- The unlock reverts what was done by the Lock operation.- It can only be called on same and unlocked by same uid. 3. UpgradeLock(X, uid): The user uid can upgrade their lock to an ancestor node.- It is only possible…
- Given a world map in the form of a Generic M-ary Tree consisting of N nodes and an array of queries[], the task is to implement the functions Lock, Unlock and Upgrade for the given tree. For each query in queries[], the functions return true when the operation is performed successfully, otherwise, it returns false. The functions are defined as: X: Name of the node in the tree and will be uniqueuid: User Id for the person who accesses node X 1. Lock(X, uid): Lock takes exclusive access to the subtree rooted.- Once Lock(X, uid) succeeds, then lock(A, any user) should fail, where A is a descendant of X.- Lock(B. any user) should fail where X is a descendant of B.- Lock operation cannot be performed on a node that is already locked. 2. Unlock(X, uid): To unlock the locked node.- The unlock reverts what was done by the Lock operation.- It can only be called on same and unlocked by same uid. 3. UpgradeLock(X, uid): The user uid can upgrade their lock to an ancestor node.- It is only possible…Given a world map in the form of a Generic M-ary Tree consisting of N nodes and an array of queries[], the task is to implement the functions Lock, Unlock and Upgrade for the given tree. For each query in queries[], the functions return true when the operation is performed successfully, otherwise, it returns false. The functions are defined as: X: Name of the node in the tree and will be uniqueuid: User Id for the person who accesses node X 1. Lock(X, uid): Lock takes exclusive access to the subtree rooted.- Once Lock(X, uid) succeeds, then lock(A, any user) should fail, where A is a descendant of X.- Lock(B. any user) should fail where X is a descendant of B.- Lock operation cannot be performed on a node that is already locked. 2. Unlock(X, uid): To unlock the locked node.- The unlock reverts what was done by the Lock operation.- It can only be called on same and unlocked by same uid. 3. UpgradeLock(X, uid): The user uid can upgrade their lock to an ancestor node.- It is only possible…Another important property of a binary search tree is being able to easily find the maximum and minimum key in the tree. Discuss a simple rule to identify these keys. What is the output of a BFS traversal in a normal binary search tree if the key insert order is "1,2,3,4,5,6,7"?
- How might storing linked list pointers in an AVL tree affect insertions, deletions, and hash table searches? In your answer, you must provide the running time in terms of the size of the hash table. Big-Ohnotation is mandatory?The order of an internal node in a B* tree index is the maximum number of children it can have. Suppose that a child pointer takes 6 bytes, the search field value takes 14 bytes., and the block size is 512 bytes. What is the order of the internal node?Question 3: a. Show the result of inserting 3, 1, 4, 6, 9, 2, 5, 7 into an initially empty binary search tree.b. Show the result of deleting the root.
- Given any set of n distinct keys (identifiers) and an arbitrary binary tree T on n nodes, use mathematical induction to show that there is a unique assignment of the keys to the nodes in T such that T becomes a binary search tree for the keys. Hint. First show that there is a unique assignment of a key to the root 1.If one were to implement a list ADT using a balanced binary search tree, how would the worst-case time complexity of such an implementation compare with that of an array-based list implementation for operations set, get, and remove? Explain your answer, but be brief.1. Suppose that we are given a binary search tree T with distinct keys. We would like to find out the kth smallest and largest keys in T. The value of k is of course smaller than the size of T. Please provide efficient approaches to find these two keys and analyze your methods.