@Raj N
It won't work for the tree like. your method would return true for the
following tree.
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On Sat, Jul 3, 2010 at 3:45 AM, Raj N <[email protected]> wrote:
> 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 <[email protected]>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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