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(cleanup/format solution code, added type signature and explanation)
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<haskell>
 
<haskell>
  +
cbalTree :: Int -> [Tree Char]
 
cbalTree 0 = [Empty]
 
cbalTree 0 = [Empty]
cbalTree n = [Branch 'x' l r | i <- [q .. q + r], l <- cbalTree i, r <- cbalTree (n - i - 1)]
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cbalTree n = let (q, r) = (n - 1) `quotRem` 2
where (q, r) = quotRem (n-1) 2
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in [Branch 'x' left right | i <- [q .. q + r],
  +
left <- cbalTree i,
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right <- cbalTree (n - i - 1)]
 
</haskell>
 
</haskell>
   
Here we use the list monad to enumerate all the trees, in a style that is more natural than standard backtracking.
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This solution uses a list comprehension to enumerate all the trees, in a style that is more natural than standard backtracking.
  +
  +
The base case is a tree of size 0, for which <tt>Empty</tt> is the only possibility. Trees of size <tt>n == 1</tt> or larger consist of a branch, having left and right subtrees with sizes that sum up to <tt>n - 1</tt>. This is accomplished by getting the quotient and remainder of <tt>(n - 1)</tt> divided by two; the remainder will be 0 if <tt>n</tt> is odd, and 1 if <tt>n</tt> is even. For <tt>n == 4</tt>, <tt>(q, r) = (1, 1)</tt>.
  +
  +
Inside the list comprehension, <tt>i</tt> varies from <tt>q</tt> to <tt>q + r</tt>. In our <tt>n == 4</tt> example, <tt>i</tt> will vary from 1 to 2. We recursively get all possible left subtrees of size <tt>[1..2]</tt>, and all right subtrees with the remaining elements.
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  +
When we recursively call <tt>cbalTree 1</tt>, <tt>q</tt> and <tt>r</tt> will both be 0, thus <tt>i</tt> will be 0, and the left subtree will simply be <tt>Empty</tt>. The same goes for the right subtree, since <tt>n - i - 1</tt> is 0. This gives back a branch with no children--a "leaf" node:
  +
  +
<haskell>
  +
> cbalTree 1
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[Branch 'x' Empty Empty]
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</haskell>
  +
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The call to <tt>cbalTree 2</tt> sets <tt>(q, r) = (0, 1)</tt>, so we'll get back a list of two possible subtrees. One has an empty left branch, the other an empty right branch:
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<haskell>
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> cbalTree 2
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[
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Branch 'x' Empty (Branch 'x' Empty Empty),
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Branch 'x' (Branch 'x' Empty Empty) Empty
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]
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</haskell>
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  +
In this way, balances trees of any size can be built recursively from smaller trees.

Revision as of 21:01, 19 July 2010

(**) Construct completely balanced binary trees

In a completely balanced binary tree, the following property holds for every node: The number of nodes in its left subtree and the number of nodes in its right subtree are almost equal, which means their difference is not greater than one.

Write a function cbal-tree to construct completely balanced binary trees for a given number of nodes. The predicate should generate all solutions via backtracking. Put the letter 'x' as information into all nodes of the tree.

cbalTree :: Int -> [Tree Char]
cbalTree 0 = [Empty]
cbalTree n = let (q, r) = (n - 1) `quotRem` 2
    in [Branch 'x' left right | i     <- [q .. q + r],
                                left  <- cbalTree i,
                                right <- cbalTree (n - i - 1)]

This solution uses a list comprehension to enumerate all the trees, in a style that is more natural than standard backtracking.

The base case is a tree of size 0, for which Empty is the only possibility. Trees of size n == 1 or larger consist of a branch, having left and right subtrees with sizes that sum up to n - 1. This is accomplished by getting the quotient and remainder of (n - 1) divided by two; the remainder will be 0 if n is odd, and 1 if n is even. For n == 4, (q, r) = (1, 1).

Inside the list comprehension, i varies from q to q + r. In our n == 4 example, i will vary from 1 to 2. We recursively get all possible left subtrees of size [1..2], and all right subtrees with the remaining elements.

When we recursively call cbalTree 1, q and r will both be 0, thus i will be 0, and the left subtree will simply be Empty. The same goes for the right subtree, since n - i - 1 is 0. This gives back a branch with no children--a "leaf" node:

> cbalTree 1
[Branch 'x' Empty Empty]

The call to cbalTree 2 sets (q, r) = (0, 1), so we'll get back a list of two possible subtrees. One has an empty left branch, the other an empty right branch:

> cbalTree 2
[
  Branch 'x' Empty (Branch 'x' Empty Empty),
  Branch 'x' (Branch 'x' Empty Empty) Empty
]

In this way, balances trees of any size can be built recursively from smaller trees.