获取给定父母的所有后代

Question

你好，我有一个类别列表：

public class Category{
   private Integer id;
   private Integer parentId;
}

如何获取给定父母的所有后代/孩子的列表？

例如：

  1 : 2
  1 : 3
  2 : 4

我得到1、2、3、4

Answer 1

您可以使用这种简单的方法（可能效率不高），该方法使用递归尝试

public List<Integer> getChildrenIds(Integer parentId, List<Category> categories){
    Set<Integer> knownParents = new HashSet<>();
    List<Integer> result = new LinkedList<>();
    for(Category category : categories){
        // if category has the parent we're looking for
        if(category.parentId.equals(parentId)){
            // then add it to our result
            result.add(category.id);
            // this check is used to get all the children recursivly
            // but should be done only once, which is why we are using an intermediate set
            if(!knownParents.contains(category.id)){
                knownParents.add(category.id);
                result.addAll(getChildrenIds(category.id, categories));
            }            
        }
    }
    return result;
}

可以这样称呼：

List<Integer> childrenIds = getChildrenIds(1, categories);

注意

当数据不一致时，这会以StackOverflowError中断。

例如当具有2个类别时，彼此引用：

Category(id: 1, parentId: 2)
Category(id: 2, parentId: 1)

Answer 2

您可以遍历所有类别的列表，并检查给定ID的 direct 个子代，请参见以下示例：

public class MainCategoryChecker {

    public static void main(String[] args) {
        // create some categories and put them in a List
        List<Category> categories = new ArrayList<Category>();

        Category one = new Category(1, 0);  // the root category has parent id = 0
        categories.add(one);
        Category two = new Category(2, 1);
        categories.add(two);
        Category three = new Category(3, 1);
        categories.add(three);
        Category four = new Category(4, 2);
        categories.add(four);

        printSubCategories(1, categories);
    }

    public static void printSubCategories(int parentId, List<Category> categories) {
        for (Category c : categories) {
            if (c.getParentId() == parentId) {
                System.out.println(parentId + " : " + c.getId());
            }
        }
    }
}