Question

In: Computer Science

You will be building 4 data structure types from scratch in C++ AND Java: -Undirected Graphs...

You will be building 4 data structure types from scratch in C++ AND Java:

-Undirected Graphs
-Directed Graphs
-Weighted Graph
-Binary Search Tree

Each of these data structures will have the following requirements:

-Undirected Graphs / Directed Graphs / Weighted Graph
-Methods:
-NumOfNodes
-NumOfEdges
-PrintNodes
-PrintEdges
-IsEqual
-AddNode
-AddEdge
-RemoveNode
-RemoveEdge
-Neighbors ( Adjacency List )
-BreadthFirstSearch
-DepthFirstSearch
-Binary Search Tree
-Methods:
-InOrderTraversal
-PostOrderTraversal
-PreOrderTraversal
-Search
-Insert
-RemoveNode
-DeleteTree
-FindMin
-FindMax
-IsEqual

Solutions

Expert Solution

Please give positive ratings for my effort. Kindly follow the comments for more. Thanks.

PROGRAM

#include <iostream>
using namespace std;

struct Node
{
   int data;
   struct Node* left, *right;
   Node(int data)
   {
       this->data = data;
       left = right = NULL;
   }
};

void printPostorder(struct Node* node)
{
   if (node == NULL)
       return;

   printPostorder(node->left);

   printPostorder(node->right);

   cout << node->data << " ";
}

void printInorder(struct Node* node)
{
   if (node == NULL)
       return;

   printInorder(node->left);

   cout << node->data << " ";

   printInorder(node->right);
}

void printPreorder(struct Node* node)
{
   if (node == NULL)
       return;

   cout << node->data << " ";

   printPreorder(node->left);

   printPreorder(node->right);
}

int main()
{
   struct Node *root = new Node(1);
   root->left = new Node(2);
   root->right   = new Node(3);
   root->left->left = new Node(4);
   root->left->right = new Node(5);

   cout << "\nPreorder traversal of binary tree is \n";
   printPreorder(root);

   cout << "\nInorder traversal of binary tree is \n";
   printInorder(root);

   cout << "\nPostorder traversal of binary tree is \n";
   printPostorder(root);

   return 0;
}

IMAGE OF PROGRAM

IMAGE OF OUTPUT


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