binary-tree

问题 How can a sorted doubly linked list be converted to a balanced binary search tree. I was thinking of doing this the same way as converting an array to a balanced BST. Find the centre and then recursively convert the left part and the right part of the DLL. For example, 1 2 3 4 5 => 1 2 (3) 4 5 => 3 / \ 2 4 / \ 1 5 This is leads to the recurrence T(n) = 2T(n/2) + O(n). O(n) is for finding the centre. The time complexity is therefore O(nlogn). I was wondering if there is an algorithm that does

问题 I have just started studying Binary Tree. Is there an algorithm to find out the binary tree structure,given the Inorder and Postorder OR Inorder and Preorder? I have been trying to do it manually,but it never comes out correct.For eg.-These two are valid Inorder and Postorder traversal of a given tree: Inorder: D B F E A G C L J H K Postorder : D F E B G L J K H C A Clearly A is the root as it is the last element in Postorder. Now looking in Inorder,the left subtree becomes: {D B F E} and

问题 How to find a loop in a binary tree? I am looking for a solution other than marking the visited nodes as visited or doing a address hashing. Any ideas? 回答1: As mentioned already: A tree does not (by definition) contain cycles (loops). To test if your directed graph contains cycles (references to nodes already added to the tree) you can iterate trough the tree and add each node to a visited-list (or the hash of it if you rather prefer) and check each new node if it is in the list. Plenty of

问题 I am studying red-black trees and I am reading the Cormen's "Introduction to Algorithms" book. Now I am trying to create red-black tree with numbers 1-10 by using the pseudo-code described in the book - RB-INSERT-FIXUP(T, z). Here is the screenshot Everything was fine until I inserted number "6" into the tree. According to pseudo-code I get the following result As you can see all red-black tree requirements met, but I am confused because I know that red-black tree should be balanced on each

问题 im coming from c++ to java and i am confused on binary trees with java. is the only way to have a Node class is to make it an inner static class? all the examples i see do this. However, the way im doing it is i have a node class and a binarytree class uses this node class. but i keep getting an error when i try inserting into the tree after the second insert. i get an exception at this line if(dataIn <= nodeIn.getLeft().getData()){ I am confused as to what i did wrong.... here is my code for

问题 I understand that the degree of a node is the number of children it has. However, how do we define the degree of a tree? 回答1: Basically The degree of the tree is the total number of it's children i-e the total number nodes that originate from it.The leaf of the tree doesnot have any child so its degree is zero. The degree of a node is the number of partitions in the subtree which has that node as the root. Nodes with degree=0 are called leaves. 回答2: In general a graph has a minimum degree and

问题 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

问题 While i was studying for midterm about binary trees, i found a statement that any arbitrary n-node binary tree can be transformed into any other n-node binary tree with at most 2*n-2 rotations. Is there any proof for that? I found some kind of proof with asymptotic notations but it was not that clear. I mean could someone explain/show why it is true? And if it says that n-node binary tree, does it include the root? 回答1: This answer is from CLRS 3rd Edition textbook question 13.2-4. Let LEFT =

问题 While i was studying for midterm about binary trees, i found a statement that any arbitrary n-node binary tree can be transformed into any other n-node binary tree with at most 2*n-2 rotations. Is there any proof for that? I found some kind of proof with asymptotic notations but it was not that clear. I mean could someone explain/show why it is true? And if it says that n-node binary tree, does it include the root? 回答1: This answer is from CLRS 3rd Edition textbook question 13.2-4. Let LEFT =

问题 While i was studying for midterm about binary trees, i found a statement that any arbitrary n-node binary tree can be transformed into any other n-node binary tree with at most 2*n-2 rotations. Is there any proof for that? I found some kind of proof with asymptotic notations but it was not that clear. I mean could someone explain/show why it is true? And if it says that n-node binary tree, does it include the root? 回答1: This answer is from CLRS 3rd Edition textbook question 13.2-4. Let LEFT =