您可以使用Java8流来实现这一点。考虑以下代码：

// Testdata - The first map
Map<String, Long> m1 = new TreeMap<>();
m1.put("A", 1L);
m1.put("B", 1L);
m1.put("C", 1L);

// Testdata - The second map
Map<String, Long> m2 = new TreeMap<>();
m2.put("C", 2L);
m2.put("D", 2L);
m2.put("E", 2L);

// Summarize using streams
final Map<String, Long> summarized =
        Stream.concat(m1.entrySet().stream(), m2.entrySet().stream())     // Stream both maps
              .collect(Collectors.groupingBy(                             // Collect the map
                          Map.Entry::getKey,                              // Group by key
                          Collectors.summingLong(Map.Entry::getValue)));  // Value is the sum

System.out.println("Summarized: " + summarized);                          // Print the output

总结 Map 按键分组，按值汇总。输出为：
总结：{a=1，b=1，c=3，d=2，e=2}
如果要将其放入函数中，只需执行以下操作：

public Map<String, Long> summarize(
        final Map<String, Long> m1, 
        final Map<String, Long> m2) {

    return Stream.concat(m1.entrySet().stream(), m2.entrySet().stream())
                 .collect(groupingBy(
                          Map.Entry::getKey,
                          summingLong(Map.Entry::getValue)));
}

要阅读有关Java8流的更多信息，请查看oracle文档。

以下函数计算总和：

public static TreeMap<String, Long> sum(TreeMap<String, Long> first,
        TreeMap<String, Long> second) {
    TreeMap<String, Long> result = new TreeMap<String, Long>(first);

    for (Entry<String, Long> e : second.entrySet()) {
        Long l = result.get(e.getKey());
        result.put(e.getKey(), e.getValue() + (l == null ? 0 : l));
    }

    return result;
}

测试代码：

TreeMap<String, Long> first = new TreeMap<String, Long>();
TreeMap<String, Long> second = new TreeMap<String, Long>();

first.put("x", 1L);
first.put("y", 5L);

second.put("x", 2L);
second.put("y", 3L);
second.put("z", 5L);

System.out.println(sum(first, second));

输出：

{x=3, y=8, z=5}

编辑
一个小的优化就是复制最大的 TreeMap 迭代最小的。这减少了查找/插入的数量。

public static TreeMap<String, Long> sum(TreeMap<String, Long> first,
        TreeMap<String, Long> second) {
    // optimization (copy the largest tree map and iterate over the
    // smallest)
    if (first.size() < second.size()) {
        TreeMap<String, Long> t = first;
        first = second;
        second = t;
    }

    TreeMap<String, Long> result = new TreeMap<String, Long>(first);

    for (Entry<String, Long> e : second.entrySet()) {
        Long l = result.get(e.getKey());
        result.put(e.getKey(), e.getValue() + (l == null ? 0 : l));
    }

    return result;
}