binary-tree

Binary Tree in Objective-C

丶灬走出姿态 提交于 2019-12-02 16:52:13
I am learning algorithms and data structures and to train I am trying to design and implement a binary tree using objective-c. So far I have the following Classes: main - for testing Node - node of tree BinaryTree - for all methods related to the tree One of the first methods in BinaryTree class I implemented is insertNode:forRoot: . - (void)insertNodeByRef:(Node **)node forRoot:(Node **)root{ if (head == NULL) { head = *node; } // Case 2 root is null so can assign the value of the node to it if (root == NULL) { root = node; } else { if (node.data > root.data) { // to the right [self

check if a tree is a binary search tree

邮差的信 提交于 2019-12-02 16:46:07
I have written the following code to check if a tree is a Binary search tree. Please help me check the code: Okay! The code is edited now. This simple solution was suggested by someone in the posts below: IsValidBST(root,-infinity,infinity); bool IsValidBST(BinaryNode node, int MIN, int MAX) { if(node == null) return true; if(node.element > MIN && node.element < MAX && IsValidBST(node.left,MIN,node.element) && IsValidBST(node.right,node.element,MAX)) return true; else return false; } A Method should only do one thing at a time. Also the way you do things are generally Weird. I will give you

B trees vs binary trees

笑着哭i 提交于 2019-12-02 16:12:53
If I am implementing in-memory(RAM) search operation with b trees, then would it be better in terms of caching or some other effects when compared with binary trees? What I know is binary search tress---O(log n) btrees ---------------O(c log n) there was a lot of discussion regarding that on various blogs. Algorithmic complexity is the same, since O(log b n) = O(c log n) = O(log n) but the constant factors, which are hidden in big-O notation, could vary noticeably, depending on implementation and hardware. B-trees were designed for platter hard disks, which have a large access time (moving the

Linked list implementation of Binary Min Heap (Having trouble with manipulation…)

拟墨画扇 提交于 2019-12-02 15:51:46
问题 So i'm trying to implement a binary min heap. I understand what the binary min heap entails in terms of it's structure and it's properties. However I'm hitting a wall when I'm trying to implement it using pointers and nodes. Im using a Node , which has right/left and pointers , int element and parent pointer . I also have a LastNode in place which points to the last node inserted. My quarrel is I dont know what to do when I insert an element, in terms of the last Node. Here is what I mean.

Find paths in a binary search tree summing to a target value

£可爱£侵袭症+ 提交于 2019-12-02 14:32:22
Given a binary search tree and a target value, find all the paths (if there exists more than one) which sum up to the target value. It can be any path in the tree. It doesn't have to be from the root. For example, in the following binary search tree: 2 / \ 1 3 when the sum should be 6, the path 1 -> 2 -> 3 should be printed. Traverse through the tree from the root and do a post-order gathering of all path sums. Use a hashtable to store the possible paths rooted at a node and going down-only. We can construct all paths going through a node from itself and its childrens' paths. Here is psuedo

Real world examples of tree structures

怎甘沉沦 提交于 2019-12-02 14:19:35
I'm looking for some examples of tree structures that are used in commercial/free software projects, modern or old. I can see examples on wikipedia, but I am looking for more concrete examples and how they're used. For example primary keys in databases are (from what I've read) stored in BST structure or a variation of the BST (feel free to correct me on this) My question isn't limited Binary Search Trees (BSTs), it can include any variation such as red-black, AVL and so on. dirkgently Is it okay if the examples are a tad bit generic i.e. relate to graphs and not necessarily to trees? If it is

Check if a binary tree is a mirror image or symmetric

谁说我不能喝 提交于 2019-12-02 13:53:18
What is the basics algorithm for testing if a tree is symmetrical. Because it is a binary tree, I would assume that it would be a recursive definition of sorts The formal question is below: A binary tree is a mirror image of itself if its left and right subtrees are identical mirror images i.e., the binary tree is symmetrical. This is best explained with a few examples. 1 / \ 2 2 TRUE 1 / \ 2 2 \ 3 FALSE 1 / \ 2 2 / \ / \ 4 3 3 4 TRUE 1 / \ 2 2 / \ / \ 3 4 3 4 FALSE 1 / \ 2 2 / \ 3 3 TRUE In a programming language of choice, define a BTree class/C struct and an associated method to check if

Root node should have an assigned node, but remains NULL. Why can't I assign a node to my root?

亡梦爱人 提交于 2019-12-02 13:28:12
I'm writing a binary tree in object-oriented format. I've had experience with binary trees before, but it's been a while since I've touched on this. My problem is that I'm unable to assign a node to my root. Every time I check in debugging mode, the root remains NULL. While this is happening, the cur node contains all the information it's assigned. I've tried making my root private and changing this->root = NULL; to root-> = NULL; . I've also tried making all of my functions public, but it didn't make a difference. I tried declaring root's children to NULL values and name to an empty string as

OpenCL - copy Tree to device memory

帅比萌擦擦* 提交于 2019-12-02 12:25:09
I'm implemented a Binary-Search-Tree in C code. Each of my tree nodes looks like this: typedef struct treeNode { int key; struct treeNode *right; struct treeNode *left; } treeNode_t; The construction of the Tree made by the Host. The query of the tree made by the device. Now, let's assumed that I'm already finished building my Tree in host memory. I'm want to copy the root of my tree to the memory of my device. Copying the root of the tree it self isn't enough. Because the right \ left child isn't located in the device memory. This is a problem. So, my question is what is the easiest way to

Rebuild a binary tree from preorder and inorder lists

穿精又带淫゛_ 提交于 2019-12-02 11:17:47
问题 Hi I'm trying to rebuild a binary tree, I almost got it, except it throws me an error and I don't know why buildTree :: (Ord a, Eq a) => [a] -> [a] -> Tree a buildTree [] [] = Empty buildTree preOrd inOrd = Node root left right where root = head preOrd left = buildTree leftPreOrd leftInOrd right = buildTree rigthPreOrd leftInOrd Just rootInd = elemIndex root inOrd leftPreOrd = tail (take (rootInd + 1) preOrd) rigthPreOrd = tail (drop rootInd preOrd) leftInOrd = take rootInd inOrd rightInord =