binary-search-tree

问题 According to Ron Wein your able to do split and concatenation of red-black tree's in O(log(n)) time. See his artikle: Efficient Implementation of Red-Black Trees with Split and Catenate Operations However I'm still not convinced that the running time of split really is true. The idea is that split uses worst-case log(n) concatenatations. These concat's is done fast as we can find the node, p, by remembering the p, from last concatenate. The problem is that concatenation starts the fix-up

问题 According to Ron Wein your able to do split and concatenation of red-black tree's in O(log(n)) time. See his artikle: Efficient Implementation of Red-Black Trees with Split and Catenate Operations However I'm still not convinced that the running time of split really is true. The idea is that split uses worst-case log(n) concatenatations. These concat's is done fast as we can find the node, p, by remembering the p, from last concatenate. The problem is that concatenation starts the fix-up

问题 I created a doubly-linked list, and the benefits of a sentinel node were clear - no null checks or special cases at list boundaries. Now I'm writing a red black tree, and trying to figure out if there is any benefit to such a concept. My implementation is based on the last two functions in this article (top down insertion/deletion). The author uses a "dummy tree root" or "head" to avoid special cases at the root for his insertion/deletion algorithms. The author's head node is scoped to the

问题 How can I create an in-order traversal for my Binary Tree class. I have looked and tried other examples out there but cant get seem to get anything to work for me. Below is what I have so far for my in-order traversal method: public void inOrder(TreeNode<T> root) { if(root != null) { inOrder(root.left()); //Visit the node by Printing the node data System.out.printf("%d ",root.value()); inOrder(root.right()); } } public static void main(String[] args) { BinaryTree<Integer> tree = new

问题 How can I create an in-order traversal for my Binary Tree class. I have looked and tried other examples out there but cant get seem to get anything to work for me. Below is what I have so far for my in-order traversal method: public void inOrder(TreeNode<T> root) { if(root != null) { inOrder(root.left()); //Visit the node by Printing the node data System.out.printf("%d ",root.value()); inOrder(root.right()); } } public static void main(String[] args) { BinaryTree<Integer> tree = new

问题 How can I create an in-order traversal for my Binary Tree class. I have looked and tried other examples out there but cant get seem to get anything to work for me. Below is what I have so far for my in-order traversal method: public void inOrder(TreeNode<T> root) { if(root != null) { inOrder(root.left()); //Visit the node by Printing the node data System.out.printf("%d ",root.value()); inOrder(root.right()); } } public static void main(String[] args) { BinaryTree<Integer> tree = new

问题 I did the LeetCode question Binary Search Tree Iterator. the following code is what I learned from others. One part I didn't understand which is cur = cur.right. Since I got the smallest value of cur.val. Why do I need to assign cur.right to cur? When I remove cur = cur.right, it said time limit exceeded. Could someone can help me to explain it? public class BSTIterator { Stack<TreeNode> stack; TreeNode cur; public BSTIterator(TreeNode root) { stack = new Stack<>(); cur = root; } /** @return

问题 I have read that it is log(n+1) <= h <= 2*log(n+1) where log is in base 2. However, when trying this out on a few known minimum heights, it does not always work out. So far I know that: For for h = 1, minimum # of nodes = 2. For h = 2, minimum nodes = 4. For h = 3, minimum nodes = 10. However these were done purely by tracing it out using the rules for red-black trees. Should I be taking note of the black-height when trying to find this or is my approach/calculations just completely wrong?

问题 I have read that it is log(n+1) <= h <= 2*log(n+1) where log is in base 2. However, when trying this out on a few known minimum heights, it does not always work out. So far I know that: For for h = 1, minimum # of nodes = 2. For h = 2, minimum nodes = 4. For h = 3, minimum nodes = 10. However these were done purely by tracing it out using the rules for red-black trees. Should I be taking note of the black-height when trying to find this or is my approach/calculations just completely wrong?

问题 When I am trying to print level Order of BST, this question prompted me. Here is a Pre-Order Sequence: 4, 1, 2, 3, 5, 6, 7, 8 In_order Sequence : 1, 2, 3, 4, 5, 6, 7, 8 A level order sequence for a BST with above pre_order and In_order is [4, 2, 6, 1, 3, 5, 7, 8] However, for the same Pre-order an In-order sequence this level order sequence seems possible. [4, 1, 5, 2, 6, 3, 7, 8] . I don't know how. I am trying to figure this out. I am unable to construct BST in paper (drawing) that