void printLevelOrder(BinaryTree *root) {
node *temp;
enqueue(root);
while((temp=dequeue()) != NULL)
{
printf("%d ",temp->data);
if (temp->left)
{
enqueue(temp->left);
}
if ( temp->right)
{
enqueue(temp->right);
}
}
}
i guess this approach is much better. :)
On Sat, Nov 19, 2011 at 9:06 PM, atul anand <atul.87fri...@gmail.com> wrote:
> @shady : i guess code is fine....it will print the node only when level=1
> , and level is in loop so for level = 1 it will print root
> when level=2 then it will print all nodes at level 2 and so on...
>
> i guess we can optimize this code......
>
>
> On Sat, Nov 19, 2011 at 3:28 PM, shady <sinv...@gmail.com> wrote:
>
>> this doesn't seem like level order printing, because you are simply
>> printing the tree starting with the children as the root node.
>>
>>
>> On Sat, Nov 19, 2011 at 12:57 PM, Ankuj Gupta <ankuj2...@gmail.com>wrote:
>>
>>> What is the time complexity of this code for Level Order Traversal.
>>>
>>> void printLevel(BinaryTree *p, int level) {
>>> if (!p) return;
>>> if (level == 1) {
>>> cout << p->data << " ";
>>> } else {
>>> printLevel(p->left, level-1);
>>> printLevel(p->right, level-1);
>>> }
>>> }
>>>
>>> void printLevelOrder(BinaryTree *root) {
>>> int height = maxHeight(root);
>>> for (int level = 1; level <= height; level++) {
>>> printLevel(root, level);
>>> cout << endl;
>>> }
>>> }
>>>
>>> My guess is NlogN if tree is balanced if not it will be N^2.
>>>
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