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 =…
To turn your C++ code to Visual C++(Windows environment) all you need to do is to change all of your std:char s and std:string s to std:wchar s and std::wstring s, because windows uses wide (‘w’) format for text in…
You are passing prefix_exp_str[1:] to both the recursive calls. Instead, the second recursive call needs to receive whatever is left of the input string after the first recursive call is finished. solved Creating an expression tree for prefix expression
In C++, using malloc the way you are using it is undefined behavior, as it does not perform object initialization, which means that it’s not calling the vector’s constructor (but there are other things that are not done, as initializing…
Here’s some code that works with recursion. It first finds all the options for the first query, and then recursively calls itself with the array of the objects with the same value for the given query, and at the deepest…
Yes, it can easily be done using recursion. The idea is to check if there is already a node in the tree under which the new node can be fallen. For example, if the new node is “1.1.2”, then we…
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;…
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…
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…
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 =>…
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 -…
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…
I guess you’re looking to organize the whole array so items are children of some states. Run the following snippet to get a JSON-serialized result to check it (click show code snippet to see the code and run it!): var…
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…
I won’t claim that I really know what you are asking, but still I’ll give it shot. To me this seems like a very basic recursion to traverse a tree structure with some weird output while you are doing it.…