# How to check if a binary tree is balanced

Question 4.1 of *Cracking the Coding Interview*:

Implement a function to check if a binary tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of any node never differ by more than one.

To implement this, we can just translate the English definition into a given programming language. Here it is in Haskell:

```
module BalancedTree where
import qualified Data.Maybe
data Tree = Leaf | Branch Tree Tree
isBalanced :: Tree -> Bool
isBalanced Leaf = True
isBalanced (Branch l r) =
isBalanced l && isBalanced r && diff (height l) (height r) <= 1
height :: Tree -> Int
height Leaf = 0
height (Branch l r) = max (height l) (height r) + 1
diff n m = abs (n-m)
```

This naive translation gives an `O(n*log(n))`

algorithm,
which is not too bad.
But it does a linear-time amount of work at each node
to calculate the heights of each subtree.
But we can reduce the work at each node to constant time,
by passing up a `height`

from the recursive calls.
This gives us an `O(n)`

algorithm:

```
module BalancedTree where
import qualified Data.Maybe
data Tree = Leaf | Branch Tree Tree
isBalanced :: Tree -> Bool
isBalanced = Data.Maybe.isJust . isBalancedWithHeight
isBalancedWithHeight :: Tree -> Maybe Int
isBalancedWithHeight Leaf = Just 0
isBalancedWithHeight (Branch l r) = do
lh <- isBalancedWithHeight l
rh <- isBalancedWithHeight r
if diff lh rh <= 1
then Just $ max lh rh + 1
else Nothing
diff n m = abs (n-m)
main :: IO ()
main = print $ all id [
isBalanced Leaf,
isBalanced (Branch Leaf Leaf),
isBalanced (Branch (Branch Leaf Leaf) Leaf),
not $ isBalanced (Branch Leaf (Branch Leaf (Branch Leaf Leaf)))
]
```

How did I work out that the naive algorithm is `O(n*log(n))`

?
I actually did it by noticing “this looks like a sorting algorithm”,
in that it makes recursive sub-calls then does a linear amount of work,
then remembering that "these sorting algorithms are `O(n*log(n))`

.
A more principled way is to write out the recurrence relation,
`T(n) = 2*T(n/2) + n`

,
then use magic master theorem.
I’m sure I’ll have to learn the master theorem properly in a future question from this book.

This page copyright James Fisher 2020. Content is not associated with my employer.