Shared ancestor between two views

Question

What is the most efficient way to find the lowest common ancestor between two UIView instances?

Short of implementing Lowest Common Ancestor, are there any UIKit APIs that can be leveraged to find it?

NSView has ancestorSharedWithView: so I suspect this might be added sooner than later to iOS.

I'm currently using this quick and dirty solution, which is inefficient if the given view isn't a sibling or direct ancestor.

- (UIView*)lyt_ancestorSharedWithView:(UIView*)aView
{
    if (aView == nil) return nil;
    if (self == aView) return self;
    if (self == aView.superview) return self;
    UIView *ancestor = [self.superview lyt_ancestorSharedWithView:aView];
    if (ancestor) return ancestor;
    return [self lyt_ancestorSharedWithView:aView.superview];
}

(for those implementing a similar method, the unit tests of the Lyt project might be helpful)

Answer 1

It's not too hard, using -isDescendantOfView:.

- (UIView *)my_ancestorSharedWithView:(UIView *)aView
{
    UIView *testView = self;
    while (testView && ![aView isDescendantOfView:testView])
    {
        testView = [testView superview];
    }
    return testView;
}

Answer 2

Swift 3:

extension UIView {
    func findCommonSuperWith(_ view:UIView) -> UIView? {

        var a:UIView? = self
        var b:UIView? = view
        var superSet = Set<UIView>()
        while a != nil || b != nil {

            if let aSuper = a {
                if !superSet.contains(aSuper) { superSet.insert(aSuper) }
                else { return aSuper }
            }
            if let bSuper = b {
                if !superSet.contains(bSuper) { superSet.insert(bSuper) }
                else { return bSuper }
            }
            a = a?.superview
            b = b?.superview
        }
        return nil

    }
}