induction – Every AVL tree can be colored to be a red-black tree

I want to prove any AVL tree can be turnt into a red-black tree by coloring nodes appropriately.
Let $h$ be the height of a subtree of an AVL tree.
It is given that such a coloring is constrained by these cases:

  1. $h$ odd $implies$ black height $=$ $frac{h}{2} + 1$, root node black
  2. $h$ even $implies$ black height $=$ $frac{h+1}{2}$, root node red

After that the root node is colored black.

I’m trying to prove this inductively. Let’s start with the base case $h=1$. Then there is only one node (the root node) and it gets colored black (using case 2) which yields a valid red-black tree.

Now suppose the statement is true for some $h geq 1$. Then for any node $u$ in the AVL tree, the height difference between their children is less than $1$. That is, for an AVL tree of height $h+1$ either both subtrees of the root node have height $h$ or one has height $h-1$.

By the induction hypothesis we know how to color the subtree of height $h$, depending on the parity of $h$. I’m unsure if I should use strong induction instead because it is not given in the hypothesis how to color a subtree of height $h-1$.

If we would know how to color both subtrees, then consider the following cases:

  1. $h+1$ is even
    • one subtree has height $h$, the other height $h-1$
    • both subtrees have height $h$
  2. $h+1$ is odd
    • one subtree has height $h$, the other height $h-1$
    • both subtrees have height $h$

For case 1.1 we would get
quad & h+1 &text{even} \
implies quad & h &text{odd} \
implies quad & text{black height} = frac{h}{2} + 1 \
implies quad & h-1 &text{even} \
implies quad & text{black height} = frac{(h-1)+1}{2} = frac{h}{2}

So their black heights differ by $1$. How would I take that into consideration?