Reverse alternate levels of a perfect binary tree

Given a complete binary tree, reverse the nodes present at alternate levels.

• Time : O(n)
• Space : O(n)

void reverseAlternate(Node *root)
{
    deque<Node*> dq;
    bool reverse = false;
    dq.push_back(root);
    
    while(!dq.empty()) {
        int sz = dq.size();
        if(reverse) {
            int l = 0, r = sz - 1;
            while(l < r) {
                swap(dq[l++]->data, dq[r--]->data);
            }
        }
        while(sz--) {
            auto f = dq.front(); dq.pop_front();
            if(f->left and f->right) {
                dq.push_back(f->left);
                dq.push_back(f->right);
            }
        }
        
        reverse = !reverse;
    }
}