@algoose
yes u r ryt
On 26 June 2010 18:01, Raj N <rajn...@gmail.com> wrote:
> 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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