# 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
begin{align*} 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} end{align*}

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