二叉树中的成本搜索操作？

Question

How much does it cost the search operation in a Binary tree? Is it O(n)?

Answer 1

        Average     Worst case
Space   O(n)        O(n)
Search  O(log n)    O(n)
Insert  O(log n)    O(n)
Delete  O(log n)    O(n)

Answer 2

Yes, it is O(n), since it is Binary Tree and NOT binary search tree.

Since it is not possible to judge to which way (Left or Right) to branch in a "Binary tree", we have to search the entire tree in the worst case.

Answer 3

Average Case for searching an element: O(log n)

Worst Case: O(n)

You can check out for balanced trees (AVL, Red Black) if you need better (logarithmic) worst case running complexities.

Answer 4

是，它将是O（n），因为在该树中没有像二进制搜索树那样的排序条件，因此您必须遍历整个树，就像数组一样

Answer 5

根据Robert Sedgewick的《算法分析入门》一书，如果此二叉树是通过大小为N的随机排列构造的，则平均成功搜索为2H_N − 3 + 2H_N / N = 2ln（N）+ O（ 1），搜索失败的平均值为2H_ {N + 1}-2 = 2ln（N）+ O（1）。