99 questions/Solutions/55
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<haskell> | <haskell> | ||
| + | cbalTree :: Int -> [Tree Char] | ||
cbalTree 0 = [Empty] | cbalTree 0 = [Empty] | ||
| - | cbalTree n = [Branch 'x' | + | 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)] | ||
</haskell> | </haskell> | ||
| - | + | 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. | ||
| + | |||
| + | 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 | ||
| + | [Branch 'x' Empty Empty] | ||
| + | </haskell> | ||
| + | |||
| + | 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: | ||
| + | |||
| + | <haskell> | ||
| + | > cbalTree 2 | ||
| + | [ | ||
| + | Branch 'x' Empty (Branch 'x' Empty Empty), | ||
| + | Branch 'x' (Branch 'x' Empty Empty) Empty | ||
| + | ] | ||
| + | </haskell> | ||
| + | |||
| + | 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.
