AVL tree insert case need a clarify - algorithm

Question

I read several online document about implementing AVL tree using balance factor(Not using a height data field in Node structure), when insert into the tree, there is one case I could not figure out.
The document is from here(also from here has similar function) , the balance factor is defined as height(right-subtree) - height(left-subtree), value in [-1,0,1].
Insert method is implemented as :
public avlitem insertAVL(avlitem root, avlitem newstudent)
{
if(root == null) {
root = newstudent;
unbalance = true;
}
else if(root.idno == newstudent.idno)
System.err.println("No duplicates are allowed");
else if (root.idno > newstudent.idno)
{
root.leftchild=insertAVL(root.leftchild, newstudent);
if (unbalance )
{
switch(root.bfactor)
{ case -1:
root = balancefromleft(root);
unbalance=false;
break;
case 0:
root.bfactor = -1;
unbalance = true;
break;
case 1:
root.bfactor = 0;
unbalance = false;
}
}
}
else
{
root.rightchild=insertAVL(root.rightchild, newstudent);
if (unbalance )
{
switch(root.bfactor)
{
case -1 :
root.bfactor = 0;
unbalance = false;
break;
case 0:
root.bfactor=1;
unbalance = true;
break;
case 1 :
root = balancefromright(root);
unbalance = false;
}
}
}
return root;
}
And insert into the left child ,we use balancefromleft to reblance the tree:
public avlitem balancefromleft(avlitem root)
{
avlitem p;
avlitem w;
p = root.leftchild;
switch(p.bfactor)
{
case -1:
root.bfactor=0;
p.bfactor = 0;
root = rotatetoright(root);
break;
case 0 : System.err.println("can't balance from the left");break;
case 1 : w = p.rightchild;
{
switch (w.bfactor)
{
case -1:root.bfactor = 1;p.bfactor = 0;break;
case 0: root.bfactor = 0; p.bfactor = 0;break; // what is this case ?
case 1: root.bfactor = 0;p.bfactor = -1;
}
w.bfactor = 0;
p = rotatetoleft(p);
root.leftchild = p;
root = rotatetoright(root);
}
}
return root;
}
In the above balancefromleft function, we have a right rotate and double rotate case.
1)The right rotate case is : insert into left child of node y.
T1, T2, T3 and T4 are subtrees.
z y
/ \ / \
y T4 Right Rotate (z) x z
/ \ - - - - - - - - -> / \ / \
x T3 T1 T2 T3 T4
/ \
T1 T2
2) The left then right rotate case: insert into right child of y.
z z w
/ \ / \ / \
y T4 Left Rotate (y) w T4 Right Rotate(z) y z
/ \ - - - - - - - - -> / \ - - - - - - - -> / \ / \
T1 w y T3 T1 T2 T3 T4
/ \ / \
T2 T3 T1 T2
My question:
In the left-then-right rotate case, what case can cause the w.bfactor = 0 ?
In my mind, this could only be possible when remove node of T4 and T4 become shorter than before.How could it be possible when insert node ?
switch (w.bfactor)
{
case -1:root.bfactor = 1;p.bfactor = 0;break;
case 0: root.bfactor = 0; p.bfactor = 0;break; // what is this case ?
case 1: root.bfactor = 0;p.bfactor = -1;
}