> In AA trees, we only allowed the right child to have the same level as its parent. We could weaken the invariant, so that either child (but not both)
> can have the same level as its parent. In other words, a 3-node is constructed by a node and either of its children. What happens if you do this?
> Is there an advantage to doing it? Is there a disadvantage? 

If we weaken the invariant, then breaking the invariant is harder.
For example, for the case on http://www.cse.chalmers.se/edu/course/DAT037_Datastrukturer/slides/7a-more-trees.pdf#page=40
, we do not need to skew, because left child can have the same level as the parent.

However, checking if the invariant is broken becomes much harder as we have to check more cases now.

In the previous assumption (with right based invariant), when inserting, we considered two cases to check if invariant is broken:

(number in parenthesis is the level of the node)

- inserted x and broke the invariant of left child of n

    n (1)    
   /      
  x (1)

- (created a 4 node) inserted x and broke invariant of righ-right child of m

          m (1)     
           \ 
            n (1)
             \
              x (1)

But now (with weakened invariant), we have to check five cases!

- inserted n or x, and broke the invariant of parent m

       m (1)
      / \ 
 (1) n   x (1)

- (created a 4 node) 

          m (1)     
           \ 
            n (1)
             \
              x (1)
	      
- (created a 4 node) 

      m (1)
       \
        n (1)
       /
      x (1)

- (created a 4 node)

      m (1)
     / 
    n (1)
     \
      x (1)
      
- (created a 4 node)

        m (1)
       / 
      n (1)
     /
    x (1)

Now, we also need to fix the invariant for each case!