不用递归的情况下将带父节点的表转换为树状结构

将当前单向节点（Menu1） 转换成 树状结构(Java 中为链表，Menu2)

Name Current Node Parent Node
A 1 0
B 2 0
C 3 1
D 4 3 
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public class Menu1 {
    private Long menuId;
    private String name;
    private Long parentId;

    public Menu1(Long menuId, String name, Long parentId) {
        this.menuId = menuId;
        this.name = name;
        this.parentId = parentId;
    }
	....    
}
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public class Menu2 {
    private Long menuId;
    private String name;
    private List<Menu2> children;

    public Menu2() {
    }

    public Menu2(Long menuId, String name, List<Menu2> children) {
        this.menuId = menuId;
        this.name = name;
        this.children = children;
    }
  ....
}

具体代码:

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static List<Menu2> start(List<Menu1> menu1List) {
    //找到Root 节点的所有children
    Menu2 menu2Root = new Menu2(0L, "ROOT", null);
    List<Menu2> topLevelCollection = menu1List.stream()
            .filter(data -> data.getParentId().equals(0L))
            .map(data->new Menu2(data.getMenuId(), data.getName(), null))
            .collect(Collectors.toList());
    //添加到 children 属性中
    menu2Root.setChildren(topLevelCollection);
    //创建队列，把每个节点添加进去
    Queue<Menu2> menu2Queue = new LinkedBlockingQueue<>(topLevelCollection);
    while (!menu2Queue.isEmpty()){
        Menu2 menu2 = menu2Queue.poll();
        List<Menu2> menu2collection = menu1List.stream()
                .filter(data -> data.getParentId().equals(menu2.getMenuId()))
                .map(data->new Menu2(data.getMenuId(), data.getName(), null))
                .collect(Collectors.toList());
        menu2.setChildren(menu2collection);
        menu2Queue.addAll(menu2collection);
    }

    return menu2Root.getChildren();
}