def to tree (obj: Union [int, List])-> Optional [Tree]:"""Return the Tree which represents.Return None if is not a valid representation of a Tree.You may not access Tree attributes directly, since they're private.This function can be implemented only using the Tree initializer.>>> t = to_tree ([1, [2, [5], [6]], [3, [7], [8, [10]]], [4, [9]]])>>> t._root1>>> t._subtrees [0]._root2>>> t._subtrees [1]._root3>>> t._subtrees [2]._root4

Answered: def to tree (obj: Union [int, List])->…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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#ifndef BT_NODE_H#define BT_NODE_H struct btNode{ int data; btNode* left; btNode* right;}; // pre: bst_root is root pointer of a binary search tree (may be 0 for// empty tree) and portArray has the base address of an array large// enough to hold all the data items in the binary search tree// post: The binary search tree has been traversed in-order and the data// values are written (as they are encountered) to portArray in// increasing positional order starting from the first elementvoid portToArrayInOrder(btNode* bst_root, int* portArray);void portToArrayInOrderAux(btNode* bst_root, int* portArray, int& portIndex); // pre: (none)// post: dynamic memory of all the nodes of the tree rooted at root has been// freed up (returned back to heap/freestore) and the tree is now empty// (root pointer contains the null address)void tree_clear(btNode*& root); // pre: (none)// post: # of nodes contained in tree rooted at root is returnedint…
Please answer the following section.
Please fill in the code gaps if possible. This problem has been giving me trouble. Any help is appreciated. public class Node { private String element; // we assume elements are character strings private Node next; /* Creates a node with the given element and next node. */ public Node(String s, Node n) { //fill in body } /* Returns the element of this node. */ public String getElement() { //fill in body } /* Returns the next node of this node. */ public Node getNext() { //fill in body } // Modifier methods: /* Sets the element of this node. */ public void setElement(String newElem) { //fill in body } /* Sets the next node of this node. */ public void setNext(Node newNext) { //fill in body } }
C++ please Please complete Programming Exercise 4, from pages 1402, from chapter 19 from your textbook. Write a function, singleParent, that returns the number of nodes in abinary tree that have only one child. Add this function to the classbinaryTreeType and create a program to test this function. (Note: Firstcreate a binary search tree.)
Implement LeafNode and InteriorNode classes for the expression tree as discussed on this page Use this template: Please don't change any function namesAdd any methods if necesssaryTODO: Remove the pass statements and implement the methods. '''class LeafNode:def __init__(self, data):self.data = datadef postfix(self):return str(self)def __str__(self):return str(self.data)def prefix(self):passdef infix(self):passdef value(self):return self.dataclass InteriorNode:def __init__(self, op, left_op, right_op):self.op = opself.left_op = left_opself.right_op = right_opdef postfix(self):return self.left_op.postfix() + " " + self.right_op.postfix() + " " + self.opdef prefix(self):passdef infix(self):passdef value(self):passif __name__ == "__main__":# TODO: (Optional) your test code here.a = LeafNode(4)b = InteriorNode('+', LeafNode(2), LeafNode(3))c = InteriorNode('*', a, b)c = InteriorNode('-', c, b)
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Create a function that uses Node * pointing to root of AVL Tree as an input and returns the height of the AVL tree in O(log(n)).

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