But there's an overall loop which branches backward to the same spot, regardless whether the iteration went left or right. At least that's predictable. If we

But there's an overall loop which branches backward to the same spot, regardless whether the iteration went left or right. At least that's predictable. If we have an instruction that will conditionally load one of two pointers from memory, the left or right traversal can be made branch-free. The only reason we need to switch between two code paths in the binary tree descent is that we have two different pieces of code for loading the left or right pointer to the different offsets used in the load instruction. If we use a two element array for left and right, the indices [0] and [1] select between them. We make the comparison condition calculate a 0 or 1 result and use it as an index.