There’s no special reason, to my knowledge – I’d guess that the reason is that for so many applications the highly-tuned dict and set implementations (which are hash tables) work well. They’re good enough in most cases. There are definitely situations where you need the performance characteristics of balanced binary search trees (like ordered traversal … Read more
Binary Search Trees A node in a binary tree is a data structure that has an element, and a reference to two other binary trees, typically called the left and right subtrees. I.e., a node presents an interface like this: Node: element (an element of some type) left (a binary tree, or NULL) right (a … Read more
Linked List: Item(1) -> Item(2) -> Item(3) -> Item(4) -> Item(5) -> Item(6) -> Item(7) Binary tree: Node(1) / Node(2) / \ / Node(3) RootNode(4) \ Node(5) \ / Node(6) \ Node(7) In a linked list, the items are linked together through a single next pointer. In a binary tree, each node can have 0, … Read more
You might want to look into the Day-Stout-Warren algorithm, which is an O(n)-time, O(1)-space algorithm for reshaping an arbitrary binary search tree into a complete binary tree. Intuitively, the algorithm works as follows: Using tree rotations, convert the tree into a degenerate linked list. By applying selective rotations to the linked list, convert the list … Read more
Here is a quick example of a binary insert: class Node: def __init__(self, val): self.l_child = None self.r_child = None self.data = val def binary_insert(root, node): if root is None: root = node else: if root.data > node.data: if root.l_child is None: root.l_child = node else: binary_insert(root.l_child, node) else: if root.r_child is None: root.r_child = … Read more
To answer your questions: What is the difference between AVL trees and splay trees? Both splay trees and AVL trees are binary search trees with excellent performance guarantees, but they differ in how they achieve those guarantee that performance. In an AVL tree, the shape of the tree is constrained at all times such that … Read more
Actually that is the mistake everybody does in an interview. Leftchild must be checked against (minLimitof node,node.value) Rightchild must be checked against (node.value,MaxLimit of node) 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; … Read more
The problem lies in your base case. “The height of a tree is the length of the path from the root to the deepest node in the tree. A (rooted) tree with only a node (the root) has a height of zero.” – Wikipedia If there is no node, you want to return -1 not … Read more
One advantage that no one else has pointed out is that binary search tree allows you to do range searches efficiently. In order to illustrate my idea, I want to make an extreme case. Say you want to get all the elements whose keys are between 0 to 5000. And actually there is only one … Read more
Summary Type BST (*) Heap Insert average log(n) 1 Insert worst log(n) log(n) or n (***) Find any worst log(n) n Find max worst 1 (**) 1 Create worst n log(n) n Delete worst log(n) log(n) All average times on this table are the same as their worst times except for Insert. *: everywhere in … Read more