获取TypeScript中递归属性的联合类型

假设我们有接口：

interface Node<C extends Node[] = any[]> {
    children: C
}

现在让我们假设我想要一个类型，它是元组中所有不同类型的并集。我可以做到：

type TupleOfNodeChildren<N extends Node> = N['children'];

type TypesOfNodeChildren<N extends Node> = TupleOfNodeChildren<N>[number];

请注意，

T22

既有Baz的直接子级Bar，又有Baz的子级Foo

我似乎无法让所有孩子都使用这种

类型；不管我怎么做，它总是抱怨循环引用。我假设您需要某种递归来获取所有子类的类型，但我不确定如何在没有TypeScript抱怨的情况下实现它

编辑：

下面是我尝试的一个例子：

type TypesOfAllChildren<N extends Node> = TypesOfNodeChildren<N> | TypesOfAllChildren<TypesOfNodeChildren<N>>;
//   ~~~~~~~~~~~~~~~~~~ Recursively references itself

TypesOfAllChildren=TypesOfNodeChildren | TypesOfAllChildren；
//~~~~~~~~~~~~~~~~~~~~~~~递归地引用自身

通过条件类型添加退出条件也不起作用：

type TypesOfAllChildren<N extends Node> = TypesOfNodeChildren<N> | (TypesOfNodeChildren<N> extends never ? never : TypesOfAllChildren<TypesOfNodeChildren<N>>);

TypesOfAllChildren=TypesOfNodeChildren |（TypesOfNodeChildren从不扩展？从不：TypesOfAllChildren）；
我解决了它。因此，我们：

interface Node<C extends Node[] = any[]> {
    children: C
}

type Foo = Node<[]>
type Bar = Node<[Foo, Foo]>
type Baz = Node<[Bar]>
type Faz = Node<[Baz]>

然后是一切

type T20 = TypesOfAllChildren<Foo> // never
type T21 = TypesOfAllChildren<Bar> // Foo
type T22 = TypesOfAllChildren<Baz> // Bar | Foo    <--- Includes types of deep children

type TypesOfAllChildren<N extends Node> = TypesOfNodeChildren<N> | TypesOfAllChildren<TypesOfNodeChildren<N>>;
//   ~~~~~~~~~~~~~~~~~~ Recursively references itself

type TypesOfAllChildren<N extends Node> = TypesOfNodeChildren<N> | (TypesOfNodeChildren<N> extends never ? never : TypesOfAllChildren<TypesOfNodeChildren<N>>);

interface Node<C extends Node[] = any[]> {
    children: C
}

type Foo = Node<[]>
type Bar = Node<[Foo, Foo]>
type Baz = Node<[Bar]>
type Faz = Node<[Baz]>

type SelfAndAllChildren<N extends Node> = N['children'][number] extends never ? N : {
    // @ts-ignore
    getSelfAndChildren: N | SelfAndAllChildren<N['children'][number]>
}[N extends never ? never : 'getSelfAndChildren']

type JustAllChildren<N extends Node> = SelfAndAllChildren<N['children'][number]>