If a node in the above binary search tree is to be located by binary tree search operation, what is the expected average number of comparisons required to locate one of the nodes in the tree chosen at random? Оз O 1.75 O2 O 2.75

A worst-case binary tree sort using a self-balancing binary search tree takes O(n log n) time. Merge sort is substantially faster.

In a binary search tree, to remove a node N that has left child C1 and right child C2, we do the following:   Group of answer choices We make C1 the left child of N’s parent and C2 the right child of N’s parent   We make C1 the right child of N’s parent and C2 the left child of N’s parent   We find the largest item L in N’s left subtree, copy the contents of L to N, and remove L   We find the smallest item S in N’s right subtree, copy the contents of S to N, and remove N   We find the largest item L in N's right subtree, copy the contents of L to N, and remove L

This algorithm find the value of the largest item in a binary search tree. What is the time complexity of this algorithm, and why? int maxValue(node *root) { int max; max = maximum(root) return max;   // if max equals -1, then the binary search tree was empty } int maximum(node *node) { if (node == NULL) { return -1; { else { int value = node->value; int left = maximum(node->left); int right = maximum(node->right); if (value < left) { value = left; } if (value < right) { value = right; } return value; } {