algorithm,data-structures,binary-tree,binary-search-tree,avl-tree

How to find the order of one given element? The order of an element is its rank on the resulting sorted sequence. If assume we store the number of children in the sub-tree, we can assign virtual weigths to edges. Any edge going left weigtht 0. Any edge (x...

To my understanding, both alternatives for rebalancing are correct as they yield search trees (left sucessors are smaller than the root and right successors are are larger than the root for every subtree) and each subtree is balanced, i.e. the heights of leaves differ by at most one.

The method returns height of the AVLNode, -1 otherwise. The line return t == null ? -1 : t.height is a ternary operator Is equivalent to if (t == null) { return -1 } else { return t.height } ...

c,if-statement,return,codeblocks,avl-tree

I'm not sure what your code is trying to do, since you do not define your node struct, however your function int sameFrequency(node *node, char* key) has an obvious bug: not all code paths return a value. Reformatting a bit for clarity, you can see that if strcmp(key, key)!=0 then...

algorithm,data-structures,runtime,time-complexity,avl-tree

First of all notice that, given a number x, you really only care about x mod m, so you might as well calculate that and use that as a key to enter x into the AVL tree, which is acting as a map with key (x mod m) and value...

Along with Cyber's answer, you also need to make changes to temp variable. template<class T> avl_node<T> *avlTree<T>::rr_rotation(avl_node<T> *parent) { avl_node<T> *temp; //rest of code } template<class T> avl_node<T> *avlTree<T>::ll_rotation(avl_node<T> *parent) { avl_node<T> *temp; // rest of code } template<class T> avl_node<T> *avlTree<T>::lr_rotation(avl_node<T> *parent) { avl_node<T> *temp; //rest of code }...

In an AVL tree, the hieghts of sub-trees of each node differs by at most 1. Hence if N_h denotes the height of AVL tree with N nodes, then: N_h >= 1 + N_h-1 + N_h-2 ( the heights of both sub-trees can differ by at most one ) >=...

For the first node you move up and mark the nodes. For the second node you move up and look if a node on the path is marked. As soon as you find a marked node you can stop. (And remove the marks by doing the first path again). If...

data-structures,tree,heap,binary-search-tree,avl-tree

A heap may have better insert and merge times. It really depends on the type of heap, but typically they are far less strict than an AVL because they don't have to worry about auto balancing after each operation. A heap merely guarantees that all nodes follow the same style...

In your insert method, you should consider adding rebalance(root); Just before the call updateHeight(root); at the end of the method. Then make sure to remove calls to rebalance before this. So, for example, change root.setLeftChild(rebalance(leftNodes)); to root.setLeftChild(leftNodes); Assuming the rest of your code works fine, this should be the fix....

This part jumps out at me: int AVLTree::GetDifference(Node *node) { int left = CheckHeight(node); int right = CheckHeight(node); int balance = left - right; return balance; } This will always result in left == right and therefore left - right == 0. You forgot to pass in the children: //...

c,performance,algorithm,hashtable,avl-tree

Below are some of the benefits of Hash structures, Fast lookup (O(1) theoretically) Efficient storage (helps to store key-value) Though these properties are beneficial but in some scenarios hash table can underperform. If you have large amount of objects then more storage space (Memory) will be required and thus can...

If a new node is created, it has only one node in its subtree(it is obvious). A number of elements in subtree of a node can change only if it either lies on the path from the root to a newly inserted/removed node or it is rotated. There are...

Make the return type std::pair<bool, Object>. if(/*the item exists in the tree*/) { return std::make_pair(false, object); } else { avltreeObject.insert(user_string); return std::make_pair(true, object); } ...

java,nullpointerexception,avl-tree

Since you are calling rangeS recursivly,root.left or/and root.right may be null. This will result in a NPE when root.key is accessed. Chech for null and return 0 prior to call rangeS. public static int rangeS(AvlNode root,int k1 ,int k2){ if(root == null) { return 0; } ... } Update: To...

performance,data-structures,tree,avl-tree,red-black-tree

AVL trees, compared to red/black trees, usually have smaller height because the AVL invariants give less room for imbalance. However, red/black trees, compared to AVL trees, have faster insertions and deletions (the fixup cost of maintaining the red/black invariants is lower than the fixup cost of maintaining the AVL invariants.)...

binary-tree,avl-tree,red-black-tree

I believe AVL Trees actually satisfy your condition. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. http://en.m.wikipedia.org/wiki/AVL_tree...

c++,performance,sorting,vector,avl-tree

Think of the way balancing works for AVL trees. It works best if the "middle values" come first. With a sorted input, you will need a lot of re-balancing, thus pre-sorting will probably do more harm than good. For example, consider the following AVL tree holding the values 1-6: 4...

The while (true) loop in insert() doesn't end where it should: } // The while (true) loop should end here, but it doesn't... w->up = p; /* 2. Balance AVL */ ...

c++,segmentation-fault,avl-tree

You have undefined behavior in your code, because local variables are not initialized, their values are indeterminate. You need to initialize the variables before you try to use them, e.g. node* root = nullptr; // or `0` or `NULL` ...

algorithm,data-structures,b-tree,avl-tree,tree-balancing

The answer is... yes and no. B-trees don't need to perform rotations because they have some slack with how many different keys they can pack into a node. As you add more and more keys into a B-tree, you can avoid the tree becoming lopsided by absorbing those keys into...

data-structures,hashtable,avl-tree

This is discussed directly in the Wikipedia article your referenced: Separate chaining with other structures Instead of a list, one can use any other data structure that supports the required operations. For example, by using a self-balancing tree, the theoretical worst-case time of common hash table operations (insertion, deletion, lookup)...

c,insert,segmentation-fault,avl-tree

This is because every time you insert a function, the following codes are executed: if(!(*tree)) { *tree = item; printf("Root: \n"); /*Every node seems to get printed here */ (*tree)->height = 0; return; } The reason is that insert() is called recursively. if(strcmp(item->key,(*tree)->key) < 0) { insert(&(*tree)->left, item); } else...

// in main root = avl.insert(root, item); // in avlTree::insert avl_node* avlTree::insert(avl_node* root, int item) { if (root == NULL) { // <- Here At Here, you are using ::root which is not initialized. At this point, ::root points to some random address, which is not what you want. Write...

As you can read on wikipedia In an AVL tree, the heights of the two child subtrees of any node differ by at most one That doesn't imply that you have a complete tree ! See examples of balances tree in the wikipedia page linked. So for any of the...

python,algorithm,recursion,avl-tree,iteration

Remember that recursive programs use call stacks. You can transform any recursive program to an iterative one by using a stack. In the following code, I use two. def isBalanced(root): nodes = [root] results = [] while nodes: node = nodes.pop() if node is None: results.append(-1) else if node ==...

What you understand about AVL Tree is wrong. While deleting a node your algorithm should satisfy the following things, 1) If x has no children, delete x. 2) If x has one child, delete x and link x's parent to x's child 3) If x has two children, -find x's...

java,object,reference,binary-tree,avl-tree

Java does not copy objects unless you explicitly ask it to. It is not C++. When you assign an object variable, what you're really doing is copying a reference. When you rotate a node, you need to update the parent node to point to its new child. The only exception...

I don't understand why you can't modify preOrder to do that - you just have to "remember" that current position an update it along the way: int findPositionPreOrder( struct node *root, int targetPos, int curPos) { if(root != NULL) { int newPos = findPositionPreOrder(root->left, targetPos, curPos); newPos++; if (newPos ==...

Your problem is that the path method returns a list of nodes. have two solutions define the 1st override method (str) in the node class. Or cross it as follows: from AVLTree20_3 import AVLTree tree = AVLTree() for i in range(1, 101): tree.insert(i) x = tree.getLeafNodes() for i in range(len(x)):...

Well the minimum height is easy, just fill each level of the tree with nodes until you run out. That height is the minimum. To find the maximum, do the same as for the minimum, but then go back one step (remove the last placed node) and see if adding...