According to me perform inorder traversal and at every point store the
current element in a temporary variable and check if the next element
obtained is greater than temp otherwise return false
int temp=-9999;
int flag=1;
void isBst(NODE *tree)
{
if (tree!=NULL)
{
isBst(tree->left);
if (temp<tree->info)
temp=tree->info;
else
{
flag=0;
return;
}
isBst(tree->right);
}
}
Please correct me if I'm wrong
On Fri, Jul 2, 2010 at 6:43 PM, sharad kumar <sharad20073...@gmail.com>wrote:
> i read that link ,i dont think that is very efficient,someone plzzz look at
> that soln n comment
> bcoz i m really confused in this isbst ques so plzzz
>
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