Question

Haskell, where it says

--your code goes here

import Debug.Trace f- search takes the following parameters 1. A function to check if the search is done - this function takes: a. the global information for the problerm b. the partial solution - the function returns a bool indicating if solution is complete 2. A function to find the valid next states to go to - this function takes: a. the global information for the problerm b. a partial solution -the funtion returns a list of valid next states 3. Global information for the problem 4. A partial solution - a partial solution is a list of states in reverse search returns a list of all full solutions reachable from the partial solution - full solutions will be in forward order search :: (Show a, Show b) => p trace "search " show --search p) False = undefined -Your code goes here search

Answer 1

The C++ function to check whether the given tree is Binary Search Tree is as follows:

bool isBST(struct node* root)
{
   static struct node *prev = NULL;

   // traverse the tree in inorder fashion and keep track of the previous node
   if (root)
   {
       if (!isBST(root->left))
           return false;

       // Allows only distinct valued nodes
       if (prev != NULL && root->data <= prev->data)
           return false;

       prev = root;

       return isBST(root->right);
   }

   return true;
}

Code to copy:

// ConsoleApplication28.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include<iostream>
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;

/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct node
{
   int data;
   struct node* left;
   struct node* right;
};

bool isBST(struct node* root)
{
   static struct node *prev = NULL;

   // traverse the tree in inorder fashion and keep track of the previous node
   if (root)
   {
       if (!isBST(root->left))
           return false;

       // Allows only distinct valued nodes
       if (prev != NULL && root->data <= prev->data)
           return false;

       prev = root;

       return isBST(root->right);
   }

   return true;
}

/*This function allocates a new node with the
given data and NULL left and right pointers. */
struct node* newNode(int data)
{
   struct node* node = (struct node*)malloc(sizeof(struct node));
   node->data = data;
   node->left = NULL;
   node->right = NULL;

   return(node);
}

/* Driver program to test above functions*/
int main()
{
   struct node *root = newNode(7);
   root->left = newNode(5);
   root->right = newNode(10);
   root->left->left = newNode(1);
   root->left->right = newNode(6);
   root->right->left = newNode(9);
   root->right->right = newNode(8);

   if (isBST(root))
       cout<<"The given tree is a BST";
   else
       cout<<"The given tree is not a BST";

   getchar();
   return 0;
}