After visiting a node, if we determined that it can possibly lead to a solution we will expend it. The node is called promising nonpromising root leaf

1. The below method is used to select a nodein map coloring problem.def get_node():mnode = "length = -1for k in cgraph.keys():if(len( cgraph[k]) > length and k not inok Color.keys());length = len(cgraph[k])mnode = kreturn mnode The method will return an empty string when A. All nodes are already assigned colorsB. There is no color remaining for a nodeC. Any two nodes have only one color left Note: Select 1 answer from 3 options.

Add a method in the BST class to return the number of thenonleaves as follows:/** Return the number of nonleaf nodes */public int getNumberofNonLeaves()

I could use help, me please:  Write a program (in main.cpp) that: Prompts the user for a filename containing node data. Outputs the minimal spanning tree for a given graph. You will need to implement the createSpanningGraph method in minimalSpanTreeType.h to create the graph and the weight matrix. Note: Files Ch20_Ex21Data.txt and Ch20_Ex4Data.txt contain node data that you may test your program with. minimalSpanTreeType.h: #ifndef H_queueADT #define H_queueADT      template <class Type>  class queueADT { public:     virtual bool isEmptyQueue() const = 0;        //Function to determine whether the queue is empty.       //Postcondition: Returns true if the queue is empty,       //               otherwise returns false.     virtual bool isFullQueue() const = 0;       //Function to determine whether the queue is full.       //Postcondition: Returns true if the queue is full,       //               otherwise returns false.     virtual void initializeQueue() = 0;       //Function…

You should create a method and call it findTotalleaves (); this method will find all of the nodes that are considered to be leaves, which are nodes that do not have any BNode dependents. The public function must not take any arguments, but the private method must accept as a parameter a BNode that represents the current root.

Please find the answer

There are an undetermined number of nodes in the fastest linkedlist.Is there one that stands out as being the longest?

package circularlinkedlist;import java.util.Iterator; public class CircularLinkedList<E> implements Iterable<E> { // Your variablesNode<E> head;Node<E> tail;int size; // BE SURE TO KEEP TRACK OF THE SIZE // implement this constructorpublic CircularLinkedList() {} // I highly recommend using this helper method// Return Node<E> found at the specified index// be sure to handle out of bounds casesprivate Node<E> getNode(int index ) { return null;} // attach a node to the end of the listpublic boolean add(E item) {this.add(size,item);return false; } // Cases to handle// out of bounds// adding to empty list// adding to front// adding to "end"// adding anywhere else// REMEMBER TO INCREMENT THE SIZEpublic void add(int index, E item){ }   // remove must handle the following cases// out of bounds// removing the only thing in the list// removing the first thing in the list (need to adjust the last thing in the list to point to the beginning)// removing the last…

JavaScript The jQuery "post" method that we used in this course to enable "round-trip" dialog with the node.js server has the form: $.post("/", x, y); Describe the type and purpose of the x and y parameters.

Can you make a test case for this to see if it works?class BSTnode:    def __init__(self, key=None, left=None, right=None, parent=None):        # Initialize a node.        self.__key = key  # value for the node        self.__left = left        self.__right = right        self.__parent = parent     def setkey(self, newkey):        # Sets the key value of the node to newkey        self.__key = newkey     def getkey(self):        # returns the key value of the node        return self.__key     def getright(self):        # returns the right child of the node        return self.__right     def setright(self, newright):        # Sets the right child to the new node object newright. Child nodes parent is also updated.        self.__right = newright        if newright is not None:            newright.__parent = self     def getleft(self):        # returns the left child of the node.        return self.__left     def setleft(self, newleft):        # Sets the left child to the new node object…