纳夫的答案还有更多细节：
对于相同数量级的n1和n2，这比O（n1 * log（n2））好
[1]用于从排序列表中创建平衡BST的算法（在Python中）：
def create_balanced_search_tree(iterator, n): if n == 0: return None n_left = n//2 n_right = n - 1 - n_left left = create_balanced_search_tree(iterator, n_left) node = iterator.next() right = create_balanced_search_tree(iterator, n_right) return {'left': left, 'node': node, 'right': right}
602392714
清零编程群