draw binary search tree

1. AVL Tree Definition The balanced binary search tree, also known as the AVL Tree (named by the two people who proposed this tree) is a highly balanced binary

Given a binary tree, determine if it is a valid binary search tree (BST).Assume a BST is defined as follows: The left subtree of a node contains only nodes with keys less than the node ' s key. The right subtree of a node contains only nodes with keys greater th

Given a non-empty binary search tree and a target value, find K values in the BST that is closest to the target.Note: Given target value is a floating point. Assume k is always valid, which is: k ≤total nodes. You is guaranteed to a unique set of K values in the BST is closest to the target. Follow up:Assume that the BST are balanced, could

stacks, queues, and lists all have their own advantages, and there are drawbacks as well. If I want an ordered array and a linked list of course this is very well implemented. Now I'm going to look for a value in these data structures. First say the array, because it is ordered by binary search quickly can be found. The efficiency of the search is still very high

Balance of Binary Search Trees-AVL Tree and tree rotation (illustration) The query efficiency of the Binary Search Tree is high only when the balance is maintained. It is not easy to ba

Binary search tree, or an empty tree, or:1) If its left subtree is not empty, then the value of all nodes on the left subtree is less than the value of its root node;2) If its right subtree is not empty, then the value of all nodes on the right subtree is greater than the value of its root node;3) The left and right su

Elements of a binary search tree (BST) is swapped by mistake.Recover the tree without changing its structure.Example 1:Input: [1,3,null,null,2] 1/3 2Output: [3,1,null,null,2] 3/1 2Example 2:Input: [3,1,4,null,null,2] 3/1 4 /2Output: [2,1,4,null,null,3] 2/1 4 /3Follow up: A solution using O (n) spac

storing the node probability of a keyword [1, n], q is an array that stores the probability of occurrence of pseudo-Keyword nodes (leaf nodes) [0, n] * n is the number of nodes * e [I, j] is the storage of Key Words Containing Ki ~ The expected cost of one query in the optimal binary book of Kj is as follows: q [n] and e [] q [0] of e [n + 1, n]. therefore, the range is [0, n + 1] * w [I, j], which stores Ki ~ Probability sum of Kj, w [I, j] = w [I,

index of the first element in the array that is greater than key (i.e. the smallest element larger than key), and so on. These, although only a little bit of change, the realization of the time really need to be more careful. The implementation of these binary search variants is shown in:luoxn28/algorithm_data_structure. 2, Binary