WAP to Check whether a Tree is a Binary Search Tree

This C Program Checks whether a Tree is a Binary Search Tree.
Here is source code of the C program to Check whether a Tree is a Binary Search Tree. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

Program

/*
 * C Program to Check whether a Tree is a Binary Search Tree 
 */
#include <stdio.h>
#include <stdlib.h>
struct node {
	int data;
	struct node* left;
	struct node* right;
}
;
static struct node *prev = NULL;
/*Function to check whether the tree is BST or not*/
int is_bst(struct node* root) {
	if (root) {
		if (!is_bst(root->left)) //moves towards the leftmost child of the tree and checks for the BST
		return 0;
		if (prev != NULL && root->data <= prev->data)
		            return 0;
		prev = root;
		return is_bst(root->right);
		//moves the corresponding right child of the tree and checks for the BST
	}
	return 1;
}
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);
}
int main() {
	/*
    The input tree is as shown below
                40
                / \
            20        60
            / \       \
        10        30      80
                          \
                            90
  */
	struct node *root = newNode(40);
	root->left        = newNode(20);
	root->right       = newNode(60);
	root->left->left  = newNode(10);
	root->left->right = newNode(30);
	root->right->right = newNode(80);
	root->right->right->right = newNode(90);
	if (is_bst(root))
	        printf("TREE 1 Is BST"); else
	        printf("TREE 1 Not a BST");
	prev = NULL;
	/*
    The input tree is as shown below
                50
                / \
            20        30
            / \      
        70        80 
        / \          \
    10     40     60
*/
	struct node *root1 = newNode(50);
	root1->left  = newNode(20);
	root1->right  = newNode(30);
	root1->left->left  = newNode(70);
	root1->left->right = newNode(80);
	root1->left->left->right = newNode(40);
	root1->left->left->leftt = newNode(90);
	if (is_bst1(root1))
	        printf("TREE 2 Is BST"); else
	        printf("TREE 2 Not a BST");
	return 0;
}

Output

TREE 1 Is BST
TREE 2 Not a BST