Haskell: tail recursion version of depth of binary tree
A partially tail recursive version would be this:
depth d Empty = d
depth d (Branch _ l Empty) = depth (d+1) l
depth d (Branch _ Empty r) = depth (d+1) r
depth d (Branch _ l r) = max (depth (d+1) l) (depth (d+1) r)
Note that tail rescursion in this case (as opposed to the more complex full case in Chris' answer) is done only to skip the incomplete branches.
But this should be enough under the assumption that the depth of your trees is at most some double digit number. In fact, if you properly balance your tree, this should be fine. If your trees, OTOH, use to degenerate into lists, then this already will help to avoid stack overflow (this is a hypothesis I haven't proved, but it is certainly true for a totally degenerated tree that has no branch with 2 non empty children.).
Tail recursion is not a virtue in and of itself. It is only then important if we do not want to explode the stack with what would be a simple loop in imperative programming languages.
to your 3., yes, e.g. by use of CPS technique (as shown in Chris's answer);
to your 4., correct.
to your 2., with lazy corecursive breadth-first tree traversal we naturally get a solution similar to Chris's last (i.e. his #5., depth-first traversal with explicated stack), even without any calls to max:
treedepth :: Tree a -> Int
treedepth tree = fst $ last queue
where
queue = (0,tree) : gen 1 queue
gen 0 p = []
gen len ((d,Empty) : p) = gen (len-1) p
gen len ((d,Branch _ l r) : p) = (d+1,l) : (d+1,r) : gen (len+1) p
Though both variants have space complexity of O(n) in the worst case, the worst cases themselves are different, and opposite to each other: the most degenerate trees are the worst case for depth-first traversal (DFT) and the best case (space-wise) for breadth-first (BFT); and similarly the most balanced trees are the best case for DFT and the worst for BFT.
1. Why isn't your function tail recursive?
For a recursive function to be tail recursive, all the recursive calls must be in tail position. A function is in tail position if it is the last thing to be called before the function returns. In your first example you have
depth (Branch _ l r) = 1 + max (depth l) (depth r)
which is equivalent to
depth (Branch _ l r) = (+) 1 (max (depth l) (depth r))
The last function called before the function returns is (+), so this is not tail recursive. In your second example you have
depthTR d (Branch _ l r) = let dl = depthTR (d+1) l
dr = depthTR (d+1) r
in max dl dr
which is equivalent to (once you've re-lambdified all the let statements) and re-arranged a bit
depthTR d (Branch _ l r) = max (depthTR (d+1) r) (depthTR (d+1) l)
Now the last function called before returning is max, which means that this is not tail recursive either.
2. How could you make it tail recursive?
You can make a tail recursive function using continuation-passing style. Instead of re-writing your function to take a state or an accumulator, you pass in a function (called the continuation) that is an instruction for what to do with the value computed -- i.e. instead of immediately returning to the caller, you pass whatever value you have computed to the continuation. It's an easy trick for turning any function into a tail-recursive function -- even functions that need to call themselves multiple times, as depth does. It looks something like this
depth t = go t id
where
go Empty k = k 0
go (Branch _ l r) k = go l $ \dl ->
go r $ \dr ->
k (1 + max dl dr)
Now you see that the last function called in go before it returns is itself go, so this function is tail recursive.
3. Is that it, then?
(NB this section draws from the answers to this previous question.)
No! This "trick" only pushes the problem back somewhere else. Instead of a non-tail recursive function that uses lots of stack space, we now have a tail-recursive function that eats thunks (unapplied functions) which could potentially be taking up a lot of space themselves. Fortunately, we don't need to work with arbitrary functions - in fact, there are only three kinds
\dl -> go r (\dr -> k (1 + max dl dr))(which uses the free variablesrandk)\dr -> k (1 + max dl dr)(with free variableskanddl)id(with no free variables)
Since there are only a finite number of functions, we can represent them as data
data Fun a = FunL (Tree a) (Fun a) -- the fields are 'r' and 'k'
| FunR Int (Fun a) -- the fields are 'dl' and 'k'
| FunId
We'll have to write a function eval as well, which tells us how to evaluate these "functions" at particular arguments. Now you can re-write the function as
depth t = go t FunId
where
go Empty k = eval k 0
go (Branch _ l r) k = go l (FunL r k)
eval (FunL r k) d = go r (FunR d k)
eval (FunR dl k) d = eval k (1 + max dl d)
eval (FunId) d = d
Note that both go and eval have calls to either go or eval in tail position -- therefore they are a pair of mutually tail recursive functions. So we've transformed the version of the function that used continuation-passing style into a function that uses data to represent continuations, and uses a pair of mutually recursive functions to interpret that data.
4. That sounds really complicated
Well, I guess it is. But wait! We can simplify it! If you look at the Fun a data type, you'll see that it's actually just a list, where each element is either a Tree a that we're going to compute the depth of, or it's an Int representing a depth that we've computed so far.
What's the benefit of noticing this? Well, this list actually represents the call stack of the chain of continuations from the previous section. Pushing a new item onto the list is pushing a new argument onto the call stack! So you could write
depth t = go t []
where
go Empty k = eval k 0
go (Branch _ l r) k = go l (Left r : k)
eval (Left r : k) d = go r (Right d : k)
eval (Right dl : k) d = eval k (1 + max dl d)
eval [] d = d
Each new argument you push onto the call stack is of type Either (Tree a) Int, and as the functions recurse, they keep pushing new arguments onto the stack, which are either new trees to be explored (whenever go is called) or the maximum depth found so far (whenever eval is called).
This call strategy represents a depth-first traversal of the tree, as you can see by the fact that the left tree is always explored first by go, while the right tree is always pushed onto the call stack to be explored later. Arguments are only ever popped off the call stack (in eval) when an Empty branch has been reached and can be discarded.
5. Alright... anything else?
Well, once you've noticed that you can turn the continuation-passing algorithm into a version that mimics the call stack and traverses the tree depth first, you might start to wonder whether there's a simpler algorithm that traverses the tree depth first, keeping track of the maximum depth encountered so far.
And indeed, there is. The trick is to keep a list of branches that you haven't yet explored, together with their depths, and keep track of the maximum depth you've seen so far. It looks like this
depth t = go 0 [(0,t)]
where
go depth [] = depth
go depth (t:ts) = case t of
(d, Empty) -> go (max depth d) ts
(d, Branch _ l r) -> go (max depth d) ((d+1,l):(d+1,r):ts)
I think that's about as simple as I can make this function within the constraints of ensuring that it's tail-recursive.
6. So that's what I should use?
To be honest, your original, non tail-recursive version is probably fine. The new versions aren't any more space efficient (they always have to store the list of trees that you're going to process next) but they do have the advantage of storing the trees to be processed next on the heap, rather than on the stack - and there's lots more space on the heap.
You might want to look at the partially tail-recursive function in Ingo's answer, which will help in the case when your trees are extremely unbalanced.