计算BST节点删除的时间复杂度

Question

I have a little difficulty in finding out the average and worst case time complexity. So I made this BST node removal with the following logic

When you remove a node in a binary search tree , there are 3 cases

1> The node to delete has no children. That's easy: just release its resources and you're done. Time complexity O(1)

2> The node has a single child node. Release the node and replace it with its child, so the child holds the removed node's place in the tree. Time complexity O(1)

3> The node has two children. Find the right-most child of node's left subtree. Assign its value  to root, and delete this child. **Here time compexity can be maximum O(N)**

To find the node to be deleted can be **maximum O(N)**

So how do you calculate the overall average and worst time complexity??

Answer 1

In this case, I believe a worst-case complexity will be sufficient to describe the situation.

In order to find the worst-case complexity, simply find the worst case scenario out of the possible three you mentioned (O(n) case). Therefore, the worst-case complexity is O(n).

In some rare cases (such as Quicksort), people choose to describe an average-case complexity as well as a worst-case complexity. In the case of Quicksort, it is because Quicksort is O(n*log(n)) in nearly all cases, and only reduces to O(n^2) in some very rare cases. Therefore, people describe both an average-case, as well as a worst-case complexity.

In the case of removing a node from a binary search tree, however, removing a leaf node (w/ no children and best case) is does not occur a lot more frequently or a lot less than the removal of a node w/ two children (unless you are developing for a special case).

Therefore, the complexity of the removal of a node from a binary search tree is O(n).

Answer 2

在删除的情况下，平均情况复杂度为O（log（n），因为要找到节点，将花费O（log（n），然后再次删除O（log（n）），因此平均值为O（log（n））+ O（log（n））= O（log（n））最坏的情况显然是O（n）有关更多详细信息，请参见http://en.wikipedia.org/wiki/Binary_search_tree#Deletion