the complexity is N^2
for each level you are traversing all the nodes below that level. for nth
level , you have to traverse all n-1 nodes.
so O(N^2)
better to use queue , we can traverse level order in O(n).
On Sat, Nov 19, 2011 at 4:58 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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