2009年2月8日 · Best case is O(log 2 n) for a perfectly balanced tree, since you cut the search space in half for every tree level. Average case is somewhere in between those two and depends entirely on the data :-)

We can say that for BST search, the worst-case running time is O(h), where h is the height of the tree. Similarly, the best-case running time is a function that maps input size to the minimum possible running time for that input size.

2013年10月31日 · A randomly-built BST with n nodes has a 2 n-1 / n! chance of being built degenerately, which is extremely rare as n gets to any reasonable size but still possible. In that case, a lookup might take Θ(n) time because the search might need to descend all the way down to the deepest leaf.

BST RUNNING TIME ANALYSIS Problem Solving with Computers-II. Big-Omega • f(n) and g(n) map positive integer inputs to positive reals. ... • WHAT are the (worst case) running times of each operation? 4. 5 BSTs of different heights are possible with the same set of keys Examples for keys: 12, 32, 41, 42, 45

Binary Search Tree - Best Time • All BST operations are O(d), where d is tree depth • minimum d is for a binary tree with N nodes › What is the best case tree? › What is the worst case tree? • So, best case running time of BST operations is O(log N) d=⎣log 2 N⎦ 2/1/2006 CSE 373 - AU 06 -- …

WHAT are the (worst case) running times of each operation? Path – a sequence of nodes and edges connecting a node with a descendant. Height of node – The height of a node is the number of edges on the longest downward path between that node and a leaf. Given a BST of height H and N nodes, what is the worst case complexity of searching for a key?

2023年5月10日 · 测试插入，BST在最坏情况下，在笔者的电脑上面已经爆炸了，跑了36mins都没有结束，基本GG，红黑树put()则用了 6176ms, delete()用了195ms。 如果将36mins换算成ms = 2160000ms，红黑树是千毫秒级别，而BST跑了2百万毫秒，依然没有结束。

• f(n) and g(n): running times of two algorithms on inputs of size n. • f(n) and g(n) map positive integer inputs to positive reals. We say f = O(g) if there is a constant c > 0 and k>0 such that f(n) ≤ c · g(n) for all n >= k. f = O(g) means that “f grows no faster than g”

2014年5月10日 · It really depends on the implementation of your BST, but if your BST holds a 'father' node, and is using it to find the successor, it will need to traverse each edge at most twice - once one you go "down", the first time to the node, and one when you go "up", back from it.

2021年3月19日 · The running times of algorithms on binary search trees depend on the shapes of the trees, which, in turn, depends on the order in which keys are inserted. It is reasonable, for many applications, to use the following simple model: We assume that the keys are (uniformly) random, or, equivalently, that they are inserted in random order.