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Abhijit Sarkar
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Dec 25, 2024
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| Original file line number | Diff line number | Diff line change |
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| package bintree | ||
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| final case class DList[A](run: List[A] => List[A]): | ||
| // O(1) | ||
| infix def ++(xs: DList[A]): DList[A] = DList(xs.run.andThen(run)) | ||
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| // O(1) | ||
| def toList: List[A] = run(List.empty[A]) | ||
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| object DList: | ||
| // O(1) | ||
| def empty[A]: DList[A] = DList(identity) | ||
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| // O(1) | ||
| def singleton[A]: (A => DList[A]) = a => DList(a :: _) | ||
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| // O(n) | ||
| def fromList[A]: (List[A] => DList[A]) = xs => DList(xs ++ _) |
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| Original file line number | Diff line number | Diff line change |
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| package bintree | ||
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| import Tree.* | ||
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| object P61: | ||
| extension [A](t: Tree[A]) | ||
| def leafCount: Int = t match | ||
| case Empty => 0 | ||
| case Node(_, Empty, Empty) => 1 | ||
| case Node(_, left, right) => left.leafCount + right.leafCount |
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| package bintree | ||
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| import Tree.* | ||
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| // P61A (*) Collect the leaves of a binary tree in a list. | ||
| // A leaf is a node with no successors. Write a method leafList to collect them in a list. | ||
| // | ||
| // scala> Node('a', Node('b'), Node('c', Node('d'), Node('e'))).leafList | ||
| // res0: List[Char] = List(b, d, e) | ||
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| object P61A: | ||
| extension [A](t: Tree[A]) | ||
| def leafList: List[A] = t match | ||
| case Empty => List.empty | ||
| case Node(value, Empty, Empty) => List(value) | ||
| case Node(_, left, right) => left.leafList ++ right.leafList |
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| Original file line number | Diff line number | Diff line change |
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| package bintree | ||
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| import Tree.* | ||
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| // P62 (*) Collect the internal nodes of a binary tree in a list. | ||
| // An internal node of a binary tree has either one or two non-empty successors. | ||
| // Write a method internalList to collect them in a list. | ||
| // | ||
| // scala> Node('a', Node('b'), Node('c', Node('d'), Node('e'))).internalList | ||
| // res0: List[Char] = List(a, c) | ||
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| object P62: | ||
| extension [A](t: Tree[A]) | ||
| def internalList: List[A] = t match | ||
| case Empty => List.empty | ||
| case Node(_, Empty, Empty) => List.empty | ||
| case Node(value, left, right) => left.internalList ++ (value :: right.internalList) |
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| @@ -0,0 +1,18 @@ | ||
| package bintree | ||
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| import Tree.* | ||
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| // P62B (*) Collect the nodes at a given level in a list. | ||
| // A node of a binary tree is at level N if the path from the root to the node has length N-1. | ||
| // The root node is at level 1. | ||
| // Write a method atLevel to collect all nodes at a given level in a list. | ||
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| // scala> Node('a', Node('b'), Node('c', Node('d'), Node('e'))).atLevel(2) | ||
| // res0: List[Char] = List(b, c) | ||
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| object P62B: | ||
| extension [A](t: Tree[A]) | ||
| def atLevel(level: Int, tree: Tree[A] = t, currLevel: Int = 1): List[A] = tree match | ||
| case Node(value, _, _) if currLevel == level => List(value) | ||
| case Node(_, left, right) => atLevel(level, left, currLevel + 1) ++ atLevel(level, right, currLevel + 1) | ||
| case Empty => List.empty |
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| package bintree | ||
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| import Tree.* | ||
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| // P63 (**) Construct a complete binary tree. | ||
| // A complete binary tree with height H is defined as follows: The levels | ||
| // 1,2,3,...,H-1 contain the maximum number of nodes (i.e. 2^(i-1) at the | ||
| // level i, note that we start counting the levels from 1 at the root). | ||
| // In level H, which may contain less than the maximum possible number of nodes, | ||
| // all the nodes are "left-adjusted". | ||
| // Write a method completeBinaryTree that takes as parameters the number of | ||
| // nodes and the value to put in each node. | ||
| // | ||
| // scala> Tree.completeBinaryTree(6, "x") | ||
| // res0: Node[String] = T(x T(x T(x . .) T(x . .)) T(x T(x . .) .)) | ||
| object P63: | ||
| def completeBinaryTree[A](n: Int, a: A, i: Int = 1): Tree[A] = | ||
| if i > n then Empty | ||
| else Node(a, completeBinaryTree(n, a, 2 * i), completeBinaryTree(n, a, 2 * i + 1)) |
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| package bintree | ||
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| import Tree.* | ||
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| // Problem 64: (**) Layout algorithm for displaying trees. | ||
| // In this layout strategy, the position of a node v is obtained by the following two rules: | ||
| // | ||
| // - x(v) is equal to the position of the node v in the inorder sequence | ||
| // - y(v) is equal to the depth of the node v in the tree | ||
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| // Write a function to annotate each node of the tree with a position, | ||
| // where (1,1) in the top left corner or the rectangle bounding the drawn tree. | ||
| // | ||
| // scala> Node('a', Node('b', End, Node('c')), Node('d')).layoutBinaryTree | ||
| // res0: PositionedNode[Char] = T[3,1](a T[1,2](b . T[2,3](c . .)) T[4,2](d . .)) | ||
| object P64: | ||
| type Pos = (Int, Int) | ||
| type AnnotatedTree[A] = Tree[(A, Pos)] | ||
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| extension [A](t: Tree[A]) | ||
| def layoutBinaryTree: AnnotatedTree[A] = | ||
| def loop[A](pos: Int, depth: Int, tree: Tree[A]): (AnnotatedTree[A], Int) = | ||
| tree match | ||
| case Empty => (Empty, 0) | ||
| case Node(value, l, r) => | ||
| val (left, lSize) = loop(pos, depth + 1, l) | ||
| val pos1 = lSize + pos + 1 | ||
| val (right, rSize) = loop(pos1, depth + 1, r) | ||
| val node = Node((value, (pos1, depth)), left, right) | ||
| (node, lSize + rSize + 1) | ||
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| loop(0, 1, t)._1 |
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