CareerCup

Your code is not returning result for all scenarios..
consider the case if only a single node tree say 10. It is BST but we are not returning anything so compiler would return an error i guess.

bool isBSTUtil(struct node* node, int min, int max) { 
	if (node==NULL) return false;

	if (node->data<min || node->data>max) return false;


	bool left = true;
	bool right = true;

	if(node->left)
		left = isBSTUtil(node->left, min, node->data); 
	
	if(node->right)
		right = isBSTUtil(node->right, node->data+1, max);

	return (left && right); 
}

if I do this in BStUtil
static int i=0;
if(root==null)
{
if(i==0) return false;
return true;
}
i++;
then it is correct ?

u can modify when calling recursively:

int isBSTUtil(struct node* node, int min, int max) { if (node==NULL) return(false);

if (node->data<min || node->data>max) return(false);


if(node->left)
isBSTUtil(node->left, min, node->data);
if(node->right)
return isBSTUtil(node->right, node->data+1, max);
}
Clasrufy if i made a mistake!

What about creating default parameter with value false ? something like:

int isBSTUtil(struct node* node, int min, int max, bool defaultReturnValue = false);

int isBSTUtil(struct node* node, int min, int max, bool returnValue) { if (node==NULL) return(returnValue); 
// false if this node violates the min/max constraint if (node->data<min || node->data>max) return(false);
// otherwise check the subtrees recursively,
// tightening the min or max constraint
return
isBSTUtil(node->left, min, node->data,true) &&
isBSTUtil(node->right, node->data+1, max,true)
);

The easiest way I can think of is indicating if is checking the root or not in the first call.

int isBSTUtil(struct node* node, int min, int max, bool isRoot)
{
if (isRoot && node==NULL)
return false;
else
return true;
}

On the first call on isBST2, isRoot is true. On the recursive calls isRoot is always going to be false.