#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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