Sorry that's shud have been node->right==node->right->left
On Sat, Jun 26, 2010 at 5:59 PM, Raj N <rajn...@gmail.com> wrote:
> #define max(x,y) ((x)>(y)?(x):(y))
>
> struct Bintree{
> int element;
> struct Bintree *left;
> struct Bintree *right;
> };
>
> typedef struct Bintree* Tree;
>
> int height(Tree T)
> {
> if(T->right==T->right->left)
> return -1;
> else
> return (1 + max(height(T->left), height(T->right)))
> }
>
> Instead of checking the condition for node==NULL check if
> node->right->left. This has got nothing to do with BST by the way..
>
>
> On Sat, Jun 26, 2010 at 3:54 PM, divya jain <sweetdivya....@gmail.com>wrote:
>
>> yes..
>>
>> i got the solution traverse till node->right!=node->right->left... at this
>> point u ll get height.. rite?
>>
>> On 26 June 2010 11:49, Raj N <rajn...@gmail.com> wrote:
>>
>>> @Divya: What will happen when say node->right when you reach the leaves ?
>>> Is it equivalent to node->next and node->left = = node->previous in the
>>> doubly linked list ?
>>>
>>>
>>> On Tue, Jun 22, 2010 at 4:44 PM, divya <sweetdivya....@gmail.com> wrote:
>>>
>>>> a bst is given whose leaf nodes having left as well as right pointers
>>>> not pointing to NULL. rather all the leaf nodes are forming a circular
>>>> doubly linked list. u have to calculate height of tree.
>>>>
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