在二叉查找树的第n个最大元件 [英] Nth largest element in a binary search tree

问题描述

如何找到一个BST的第N个节点？

How to find the Nth largest node in a BST?

我总是一个计数变量，而这样做在BST中序遍历？返回元素时计数= N ???

Do I keep a count variable while doing In Order Traversal of a BST? Return the element when the count = N???

推荐答案

请参阅我的回答<一href="http://stackoverflow.com/questions/2329171/find-kth-smallest-element-in-a-binary-search-tree-in-optimum-way/2329236#2329236">here.您可以在做到这一点O（log n）的平均，其中n =节点数。最坏的情况是仍然 O（N）如果树不均衡（总是 O（log n）的如果不过是平衡的）。为了遍历总是 O（N）但是。

See my answer here. You can do this in O(log n) on average where n = number of nodes. Worst case is still O(n) IF the tree isn't balanced (always O(log n) if it is balanced however). In order traversal is always O(n) however.

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