Consider a traversal of a binary tree. Suppose that visiting a node means to simply display the data in the node. What are the results of each of the following traversals of the tree in the following figures according to: a. Pre-order technique b. Post-order technique c. In-order technique A B D E F G H

Given a binary tree, let an H-node be defined as a non-leaf node in the tree whose value is greater than or equal to its children nodes (1 or 2 children). Write a function, countHNodes(), that returns the number of H-nodes in a binary tree (pointed by p) using recursion.

The BST remove algorithm traverses the tree from the root to find the node to remove. When the node being removed has 2 children, the node's successor is found and a recursive call is made. One node is visited per level, and in the worst-case scenario, the tree is traversed twice from the root to a leaf. A BST with N nodes has at least log2N levels and at most N levels. Therefore, the runtime complexity of removal is best case O(logN) and worst case O(N). Two pointers are used to traverse the tree during removal. When the node being removed has 2 children, a third pointer and a copy of one node's data are also used, and one recursive call is made. Thus, the space complexity of removal is always O(1)."   I have to explain this clearly!  and the advantages of the BST algorithim