Typescript allow/return type and its extended childrens

Question

export interface IArrayToTreeInput {}

export function convertArrayToTree(input: IArrayToTreeInput[]) {
  const roots: IArrayToTreeInput[] = [];
  return roots;
}

export interface ITree extends IArrayToTreeInput {}

let tree: ITree[] = [];
tree  = convertArrayToTree(tree);

Last line will throw error. as ITree has more values than IArrayToTreeInput. Now I can solve it by

1. allowing union type (ITree | IArrayToTreeInput) or
2. specifying as ITree (convertArrayToTree(tree) as ITree).

But is there a way where the function can return extended type without either of the above 2

Answer 1

等待大神答复