二叉搜索树的时间效率

Question

for the time efficiency of inserting into binary search tree,

I know that the best/average case of insertion is O(log n), where as the worst case is O(N).

What I'm wondering is if there is any way to ensure that we will always have best/average case when inserting besides implementing an AVL (Balanced BST)?

Thanks!

Answer 1

There is no guaranteed log n complexity without balancing a binary search tree. While searching/inserting/deleting, you have to navigate through the tree in order to position yourself at the right place and perform the operation. The key question is - what is the number of steps needed to get at the right position? If BST is balanced, you can expect on average 2^(i-1) nodes at the level i . This further means, if the tree has k levels ( k is called the height of tree), the expected number of nodes in the tree is 1 + 2 + 4 + .. + 2^(k-1) = 2^k - 1 = n , which gives k = log n , and that is the average number of steps needed to navigate from the root to the leaf.

Having said that, there are various implementations of balanced BST. You mentioned AVL, the other very popular is red-black tree , which is used eg in C++ for implementing std::map or in Java for implementing TreeMap .

The worst case, O(n) , can happen when you don't balance BST and your tree degenerates into a linked list. It is clear that in order to position at the end of the list (which is a worst case), you have to iterate through the whole list, and this requires n steps.