for loop will terminate at root only
Best Regards
Ashish Goel
"Think positive and find fuel in failure"
+919985813081
+919966006652
On Fri, Jul 9, 2010 at 10:42 PM, Balaji <balaji.subraman...@gmail.com>wrote:
> Here is one other way:
>
> Planning to do a level order traversal using a Queue and a small marker
> field whose storage in the queue is proportional to the height of the tree.
>
> Here is a psuedo code for it..
> // Returns the height of the binary tree as int.
> int find_bintree_height(root) {
> unsigned height = 0;
> if (root) {
> Queue::Push(root);
> Queue::Push("#"); // where "#" is used as marker to indicate
> level..
> }
> bool is_leaf_avail = false; // a flag to indicate whether end of tree
> is reached.
> for (;!Queue::isEmpty();) {
> ele = Queue::Pop();
> if (ele == "#") {
> if (is_leaf_avail)
> Queue::Push("#") //push it till leaf is not reached..
> is_leaf_avail = false; // reset it..
> height++;
> continue;
> } // end of outer if
> if (ele->left || ele->right)
> is_leaf_avail = true;
> if (ele->left) Queue::Push(ele->left);
> if (ele->right) Queue::Push(ele->right);
> }
> return (height)
> }
>
> This is a hasty psuedo C++ code.. Seems to be fine from cursory look.. Let
> me know
>
> Thanks,
> Balaji
>
>
> On Thu, Jul 8, 2010 at 8:48 AM, jalaj jaiswal
> <jalaj.jaiswa...@gmail.com>wrote:
>
>> @ above he asked iterative .
>> .. use a stack to eliminate recursion ...
>>
>>
>> On Thu, Jul 8, 2010 at 9:00 PM, Anand <anandut2...@gmail.com> wrote:
>>
>>> height(struct *node)
>>> {
>>> int left_height;
>>> int right_height;
>>> if(node == NULL)
>>> return 0;
>>> else
>>> left_height = height(node->left);
>>> right_height = height(node->right);
>>> return (1+ max(left_height, right_height));
>>>
>>>
>>> }
>>>
>>> On Thu, Jul 8, 2010 at 2:46 AM, sharad <sharad20073...@gmail.com> wrote:
>>>
>>>> write algo to find hieght of BINARY tree ITERATIVELY
>>>>
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>>
>>
>>
>> --
>> With Regards,
>> Jalaj Jaiswal
>> +919026283397
>> B.TECH IT
>> IIIT ALLAHABAD
>>
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