binary-search-tree

I want to convert a sorted integer array into a binary search tree. I believe I understand how to do this. I have posted my pseudo-code below. What I cannot picture is how the recursion actually works. So if my array is 1, 2, 3, 4, 5... I first make 3 the root of my BST. Then I make 2 the left-child node of 3. Then do I make 1 the left-child node of 2, and come back to 2? I don't get how the recursion steps through the whole process... Thanks in advance, and apologies if this question is poorly explained. I don't want explicit code, but I would really appreciate if someone could help me go

The rules for a Red-Black Tree: Every node is either red or black. The root is black. Every leaf (NIL) is black. If a node is red, then both its children are black. For each node, all simple paths from the node to descendant leaves contain the same number of black nodes. Rule 4 mentions that red nodes need both black childs but what if there is just one child to begin with? Is there an argument to prove or disprove this? No,a red node cannot have one child,consider the following cases:- 1.If the single child it has is red...this cannot happen because no two consecutive nodes can be red. 2.If

I need to create a recursive method that takes as a parameter the root node of a binary search tree. This recursive method will then return the int value of the total number of nodes in the entire binary search tree. This is what I have so far: public class BinarySearchTree<E> extends AbstractSet<E> { protected Entry<E> root; //called by the main method public int nodes() { return nodes(root); } //nodes() will count and return the nodes in the binary search tree private int nodes(Entry<E> current) { if(current.element != null) { if(current.left == null && current.right == null) { if(current

Pretty easy question: Recursively how can I create an array of a binary search tree (in order) which uses this constructor: public class OrderedSet<E extends Comparable<E>> { private class TreeNode { private E data; private TreeNode left, right; public TreeNode(E el) { data = el; left = null; right = null; } } private TreeNode root; public int size = 0; public OrderedSet() { root = null; } In-Order means you first have to traverse the left part of the tree, so: TreeNode tree // this is your tree you want to traverse E[] array = new E[tree.size]; // the arrays length must be equivalent to the

I have a list of tuples like list_of_tuples = [(number, name, id, parent_id), (number, name, id, parent_id), ] I am trying to sort it into an ordered structure like: { parent: [(id, name), (id, name)], parent: {parent: [(id, name)] { So, any node could have a parent and/or children I tried with: tree = defaultdict(lambda: [None, ()]) ancestors = set([item[3] for item in list_of_tuples]) for items in list_of_tuples: children_root = {} descendants = [] number, name, id, parent = items if parent is None: tree[id] = [(id, name)] elif parent: if parent not in tree.keys(): node = tree.get(parent)

I'm wondering whether there is a mapping between a sorted array (e.g., [1, 2, 3, 4, 5, 6]) and the representation that one obtains when one constructs a complete binary search tree from this sorted array, and expresses said binary search tree as an array (e.g., [4, 2, 6, 1, 3, 5], see graphic below)? 4 2 6 1 3 5 Here's some more context: It is well known that one can take a sorted array and construct a complete binary search tree from it (there is a unique representation). A recursive algorithm is: find the appropriate mid (this is actually quite tricky), treat it as the root, then recurse on