tree

[Solved] Build hierarchy type presentation of data in Excel

by

Try this, it makes use of a temporary PivotTable… Option Explicit Sub TestMakeTree() Dim wsData As Excel.Worksheet Set wsData = ThisWorkbook.Worksheets.Item(“Sheet1”) Dim rngData As Excel.Range Set rngData = wsData.Range(“Data”) ‘<—————– this differs for me Dim vTree As Variant vTree = MakeTreeUsingPivotTable(ThisWorkbook, rngData) ‘* print it out next to data, you’d choose your own destination Dim … Read more

[Solved] get tree structure of a directory with its subfolders and files using C#.net in windows application [closed]

by

Method 1 HierarchicalItem holds the information we need about an item in the folder. An item can be a folder or file. class HierarchicalItem { public string Name; public int Deepth; public HierarchicalItem(string name, int deepth) { this.Name = name; this.Deepth = deepth; } } SearchDirectory is the recursive function to convert and flatten the … Read more

[Solved] JS: Convert dot string array to object tree [duplicate]

by

You could split the strings and use the parts as path to the nested array. var array = [‘catalog.product.name’, ‘catalog.product.description’, ‘catalog.product.status’, ‘catalog.product_attribute_set.name’, ‘catalog.product_attribute_set.type’], result = []; array.forEach(s => s .split(‘.’) .reduce((object, value) => { var item = (object.children = object.children || []).find(q => q.value === value); if (!item) object.children.push(item = { value, label: value }) … Read more

[Solved] Recursive and non-recursive traversal of three degree tree

by

You’re incorrectly printing the node’s value twice. You don’t need to check node->mid, as this is checked inside inorder(node->mid);. inorder(node) { if (node) { inorder(node->left); print(“%d “, node->value); inorder(node->mid); inorder(node->right); } } 0 solved Recursive and non-recursive traversal of three degree tree

[Solved] binary tree creation in SCALA

by

You probably want something like this: trait BinTree[+A] case object Leaf extends BinTree[Nothing] case class Branch[+A](node: A, left: BinTree[A], right: BinTree[A]) extends BinTree[A] def buildTree[A](list: List[A]): BinTree[A] = list match { case Nil => Leaf case x :: xs => val (left, right) = xs.splitAt(xs.length/2) Branch(x, buildTree(left), buildTree(right)) } But you really need to get … Read more

[Solved] Generate Tree from flat Array javascript

by

You could use the level property for indicating the nested position in a helper array. Then iterate the data and build children nodes if necessary. function getTree(array) { var levels = [{}]; array.forEach(function (a) { levels.length = a.level; levels[a.level – 1].nodes = levels[a.level – 1].nodes || []; levels[a.level – 1].nodes.push(a); levels[a.level] = a; }); return … Read more

[Solved] Function to turn a Binary Tree into the sums of all nodes below each node; segmentation fault if I don’t create a new binary tree [closed]

by

It is a nuisance when we have to create an MCVE from fragments of a program. However, it’s doable — it’s a pain, but doable. Here’s my variation on your code. Of the 115 lines shown, 35 are what you wrote in the question — roughly. So, I had to create twice as much code … Read more

[Solved] Searching for an array key branch inside a larger array tree – PHP

by

I think (and you have confirmed) that you can represent your data as an XML structure. I this case, you can use XPath to find any branch in your data. Here is sample XML data inferred from your question: <?xml version=”1.0″ encoding=”utf-8″ ?> <stmt_echo> <subnodes> <exprs> <expr_constfetch> <subnodes> <name> <name> <subnodes> <parts> <part>true</part> <!– more … Read more