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Question: In C++, implement the cpp file by following the instructions given in the comments of the code #ifndef BST_H #define BST_H #include // We

Question: In C++, implement the cpp file by following the instructions given in the comments of the code

#ifndef BST_H #define BST_H

#include

// We will be storing Pokemon in the nodes of our BST struct Pokemon { int number; std::string name; std::string type; };

std::string to_string(const Pokemon& p);

class BST { public: BST(); // insert void catchPokemon(const Pokemon& p); // delete void releasePokemon(int key); // search const Pokemon* searchForPokemon(int key) const; // in-order traversal and pretty printing std::string orderedListOfCaughtPokemon() const;

// This has been implemented for you for debugging purposes std::string toGraphviz() const;

private: struct BSTNode { int key; Pokemon data; BSTNode* left; BSTNode* right; BSTNode* parent; };

// in-order traversal recursive helper std::string inOrder(BSTNode* root) const; // search helper function BSTNode* search(int key) const; // remove helper function void remove(BSTNode* key); // predecessor function, used in the remove function

BSTNode* predecessor(BSTNode* n) const;

// pointer to the root node BSTNode* root; };

#endif /* BST_H */

// cpp file

#include "bst.h"

#include #include #include #include using namespace std;

std::string to_string(const Pokemon& p) { return to_string(p.number) + ": " + p.name + " (" + p.type + ")"; }

// constructor // F?X??: make the tree start out empty BST::BST() {}

// This is the insert function // F?X??: add a new BSTNode (created on the heap) into the current tree // the key of the new node is p.number, and the data is p void BST::catchPokemon(const Pokemon& p) { // emtpy tree case: make this pokemon the root and you're done

// Otherwise, the root is not null and we need to search-- // Search for the node to make sure it's not already there. // If it's already there, just return from this function without // doing anything.

// If we got this far, add a new Node in the correct location // Remember to set the parent, left, and right properly }

// F?X??: search for a node by key // return a pointer if you find it // otherwise return nullptr if the key doesn't exist in the tree BST::BSTNode* BST::search(int key) const { return nullptr; }

// call the search() helper function to do the heavy lifting const Pokemon* BST::searchForPokemon(int key) const { BSTNode* n = search(key); if (n == nullptr) return nullptr; else return &n->data; }

std::string BST::orderedListOfCaughtPokemon() const { return inOrder(root); }\

std::string BST::inOrder(BSTNode* root) const { if (root == nullptr) return ""; else return inOrder(root->left) + to_string(root->data) + ' ' + inOrder(root->right); }

// F?X??: return the predecessor of a given node // return nullptr if there is no predecessor BST::BSTNode* BST::predecessor(BSTNode* n) const { // if n has a left node, go left and then right as far as you can

// otherwise, if n has no left pointer, one of the parents is the predecessor

// SPECIAL CASE: Return nullptr if there is no predecessor return nullptr; }

// This is the "remove" function that the user gets to see/use // F?X??: use the search and remove functions to delete a key if it exists void BST::releasePokemon(int key) { // first we need to search to find the node

// case 0: node doesn't exist; do nothing

// if the node does exist, call the remove method on it }

// F?X??: Implement this method // deletes a given node from the tree--remember to implement every case! void BST::remove(BSTNode* n) { // case 1: node is a leaf; just delete it and make the parent forget

// case 2: node has one child; replace the parent's child with that one

// case 3: node has two children; now we need to use the predecessor

// remember to actually delete the node in cases 1 and 2! }

// use this method to help you debug // paste the output into webgraphviz.com to visualize your tree // you can call this method in GDB with "call puts(b.toGraphviz())" std::string BST::toGraphviz() const { if (root == nullptr) { return "The root is null--there is no tree to draw!"; }

string out = "Paste everything between the ===== lines into " " webgraphviz.com " "==================================== " "digraph BST { " " node [fontname=\"Arial\" ]; ";

vector nodes; queue q; q.push(root); while (!q.empty()) { BSTNode* front = q.front(); q.pop(); nodes.push_back(front); if (front->left != nullptr) q.push(front->left); if (front->right != nullptr) q.push(front->right); }

list nodeDefinitions; for (BSTNode* n : nodes) { stringstream ss; ss parent != nullptr) { nodeInfo = "\ (parent is " + to_string(n->parent->key) + ")"; }

nodeDefinitions.push_back(nodeName + " [ label = \"" + to_string(n->key) + ": " + n->data.name + nodeInfo + "\" ];"); }

vector edges; for (BSTNode* n : nodes) {

stringstream ss; ss

string leftNode; if (n->left != nullptr) { ss.str(""); ss left; leftNode = "n" + ss.str(); } else { // make a null node leftNode = "nullleft" + nodeName; // this node needs to appear before the right child // just put it at the front

if (n->right != nullptr) { // we've already pushed the right node to the nodeDefinitions vector // for the correct order this left node needs to appear before it nodeDefinitions.push_front(leftNode + " [ shape = point ];"); } else { nodeDefinitions.push_back(leftNode + " [ shape = point ];"); } }

string rightNode; if (n->right != nullptr) { ss.str(""); ss right; rightNode = "n" + ss.str(); } else { // make a null node rightNode = "nullright" + nodeName; nodeDefinitions.push_back(rightNode + " [ shape = point ];"); }

// edges.push_back(nodeName + " -> { " + leftNode + " " + rightNode + " // };"); edges.push_back(nodeName + " -> " + leftNode + " [ label = \"left\"] ;"); edges.push_back(nodeName + " -> " + rightNode + " [ label = \"right\"] ;"); }

for (const string& s : nodeDefinitions) { out += " " + s + " "; } out += " "; for (const string& s : edges) { out += " " + s + " "; } out += "} " "==================================== ";

return out;

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QUESTION 4 Identify what relevant conclusion or conclusions could be made from the given premises. Every computer science major plays PC games. Vin Diesel plays PC games. Prince Harry doesn't play PC games. Mark Zuckerberg is a computer science major. Select ALL that applies. Prince Harry is not a computer science major. Vin Diesel is a computer science major Prince Harry doesn't play PC games or has a degree in computer science. Vin Diesel is not a computer science major. Mark Zuckerberg and Vin Diesel play PC games.5:35 PM 44% Done Edit = 10 Ji 1.00 ju Zajall Jijuall lis ale P Disciplines that contribute to the technical :approach to information systems include a. economics, sociology, and psychology O b. computer science, engineering, and O .networking c. operations research, management O .science, and computer science d. management science, computer science, O and engineering e. engineering, utilization management, and O computer science

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