@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 <[email protected]> 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 <[email protected]> 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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