消息编辑(完全重写)当顺序无关紧要时，就是这样。两者都应该给出Log(n) -看看其中一个是否比另一个快5%以上是有用的。HashSet可以在循环中给出O(1)测试，应该可以揭示它是否正确。

当然，HashSet实现要快得多——开销更少，因为没有排序。http://java.sun.com/docs/books/tutorial/collections/implementations/set.html提供了Java中各种Set实现的很好的分析。

这里的讨论还指出了一种有趣的“中间地带”方法来解决树与哈希的问题。Java提供了一个LinkedHashSet，它是一个HashSet，其中运行着一个“面向插入”的链表，也就是说，链表中的最后一个元素也是最近插入到哈希中的元素。这允许您避免无序散列的无序性，而不会增加TreeSet的成本。

HashSet比TreeSet快得多(对于添加、删除和包含等大多数操作，HashSet是常量时间，而不是日志时间)，但不像TreeSet那样提供排序保证。

HashSet

该类为基本操作(添加、删除、包含和大小)提供恒定的时间性能。 它不能保证元素的顺序随时间保持不变 迭代性能取决于初始容量和HashSet的负载因子。 接受默认的负载因子是相当安全的，但您可能希望指定的初始容量大约是您期望该集增长的两倍。

TreeSet

保证基本操作(添加、删除和包含)的时间成本为log(n) 确保set的元素将被排序(升序，自然或由你通过它的构造函数指定)(实现SortedSet) 不为迭代性能提供任何调优参数 提供了一些方便的方法来处理有序集，如first()， last()， headSet()和tailSet()等

重要的几点:

Both guarantee duplicate-free collection of elements It is generally faster to add elements to the HashSet and then convert the collection to a TreeSet for a duplicate-free sorted traversal. None of these implementations are synchronized. That is if multiple threads access a set concurrently, and at least one of the threads modifies the set, it must be synchronized externally. LinkedHashSet is in some sense intermediate between HashSet and TreeSet. Implemented as a hash table with a linked list running through it, however,it provides insertion-ordered iteration which is not same as sorted traversal guaranteed by TreeSet.

因此，使用方法的选择完全取决于您的需要，但我认为，即使您需要一个有序的集合，那么您仍然应该使用HashSet来创建Set，然后将其转换为TreeSet。

例如:SortedSet<String> s = new TreeSet<String>(hashSet);

A lot of answers have been given, based on technical considerations, especially around performance. According to me, choice between TreeSet and HashSet matters. But I would rather say the choice should be driven by conceptual considerations first. If, for the objects your need to manipulate, a natural ordering does not make sense, then do not use TreeSet. It is a sorted set, since it implements SortedSet. So it means you need to override function compareTo, which should be consistent with what returns function equals. For example if you have a set of objects of a class called Student, then I do not think a TreeSet would make sense, since there is no natural ordering between students. You can order them by their average grade, okay, but this is not a "natural ordering". Function compareTo would return 0 not only when two objects represent the same student, but also when two different students have the same grade. For the second case, equals would return false (unless you decide to make the latter return true when two different students have the same grade, which would make equals function have a misleading meaning, not to say a wrong meaning.) Please note this consistency between equals and compareTo is optional, but strongly recommended. Otherwise the contract of interface Set is broken, making your code misleading to other people, thus also possibly leading to unexpected behavior.

这个链接可能是关于这个问题的一个很好的信息来源。

import java.util.HashSet;
import java.util.Set;
import java.util.TreeSet;

public class HashTreeSetCompare {

    //It is generally faster to add elements to the HashSet and then
    //convert the collection to a TreeSet for a duplicate-free sorted
    //Traversal.

    //really? 
    O(Hash + tree set) > O(tree set) ??
    Really???? Why?



    public static void main(String args[]) {

        int size = 80000;
        useHashThenTreeSet(size);
        useTreeSetOnly(size);

    }

    private static void useTreeSetOnly(int size) {

        System.out.println("useTreeSetOnly: ");
        long start = System.currentTimeMillis();
        Set<String> sortedSet = new TreeSet<String>();

        for (int i = 0; i < size; i++) {
            sortedSet.add(i + "");
        }

        //System.out.println(sortedSet);
        long end = System.currentTimeMillis();

        System.out.println("useTreeSetOnly: " + (end - start));
    }

    private static void useHashThenTreeSet(int size) {

        System.out.println("useHashThenTreeSet: ");
        long start = System.currentTimeMillis();
        Set<String> set = new HashSet<String>();

        for (int i = 0; i < size; i++) {
            set.add(i + "");
        }

        Set<String> sortedSet = new TreeSet<String>(set);
        //System.out.println(sortedSet);
        long end = System.currentTimeMillis();

        System.out.println("useHashThenTreeSet: " + (end - start));
    }
}

基于@shevchyk在地图上可爱的视觉回答，以下是我的看法:

╔══════════════╦═════════════════════╦═══════════════════╦═════════════════════╗
║   Property   ║       HashSet       ║      TreeSet      ║     LinkedHashSet   ║
╠══════════════╬═════════════════════╬═══════════════════╬═════════════════════╣
║              ║  no guarantee order ║ sorted according  ║                     ║
║   Order      ║ will remain constant║ to the natural    ║    insertion-order  ║
║              ║      over time      ║    ordering       ║                     ║
╠══════════════╬═════════════════════╬═══════════════════╬═════════════════════╣
║ Add/remove   ║        O(1)         ║     O(log(n))     ║        O(1)         ║
╠══════════════╬═════════════════════╬═══════════════════╬═════════════════════╣
║              ║                     ║   NavigableSet    ║                     ║
║  Interfaces  ║         Set         ║       Set         ║         Set         ║
║              ║                     ║    SortedSet      ║                     ║
╠══════════════╬═════════════════════╬═══════════════════╬═════════════════════╣
║              ║                     ║    not allowed    ║                     ║
║  Null values ║       allowed       ║ 1st element only  ║      allowed        ║
║              ║                     ║     in Java 7     ║                     ║
╠══════════════╬═════════════════════╩═══════════════════╩═════════════════════╣
║              ║   Fail-fast behavior of an iterator cannot be guaranteed      ║
║   Fail-fast  ║ impossible to make any hard guarantees in the presence of     ║
║   behavior   ║           unsynchronized concurrent modification              ║
╠══════════════╬═══════════════════════════════════════════════════════════════╣
║      Is      ║                                                               ║
║ synchronized ║              implementation is not synchronized               ║
╚══════════════╩═══════════════════════════════════════════════════════════════╝