Yes, you are right.
How about this:
bool isBST(root *t, int min=minint, int max=maxint)
{
return !t || ((t->value >= min) && (t->value <= max) &&
isBst(t->left, min, t->value) && isBst(t->right, t->value,
max));
}
On Nov 5, 11:04 pm, atul anand <atul.87fri...@gmail.com> wrote:
> @Don : your algo wont work for following tree :-
>
> 30
> / \
> 20 31
> / \
> 9 41
>
> above tree is not a BST bcozz here 41 should lie on the right side of the
> 30....but it is not.
> so we need to keep track of max and min as we move left or right part of
> the tree.and each node should lie b/w that min and max range.
> for more details :http://www.geeksforgeeks.org/archives/3042
>
> @shady : yes correct..you can do in that way.
>
>
>
>
>
>
>
> On Tue, Nov 6, 2012 at 1:18 AM, Don <dondod...@gmail.com> wrote:
> > That would work. But a simpler approach is:
>
> > bool isBinTree(root *t)
> > {
> > return (!t) || ((!t->left || (t->value > t->left->value)) &&
> > (!t->right || (t->value < t->right->value)) &&
> > isBinTree(t->left) && isBinTree(t->right));
> > }
>
> > On Nov 5, 2:04 pm, shady <sinv...@gmail.com> wrote:
> > > Hi,
> > > Can we check this by just doing an inorder traversal, and then checking
> > if
> > > it is in increasing order or not ?
>
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