red-black-tree | 易学教程

Is there something like mulitiSet in JavaScript?

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问题 I know that JavaScript now has sets, but I wonder if there is something to realize the function of multiSet , or if there is some framework that has the functions of multiset which I really need a lot. Or I have to code it by myself to do the research of Red-Black Tree ? 回答1: There are no built-in multiset structure, but there are some libraries that have such: mnemonist → MultiSet TSTL → TreeMultiSet Feel free to add your favorite library in this question. 来源： https://stackoverflow.com

Red-Black Tree Delete Fixup in CLRS Second Edition, in Clojure

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问题 I am implementing red-black tree deletion for interval trees following CLRS 2nd edition, fourth printing, pg 288-9. Summary of bug: RB-Delete-Fixup If x and w are the sentinel nodes, which is a possible consequence of RB-Delete, then the evaluation of color(left(w)) resp. color(right(w)) in RB-Delete-Fixup suffers a null pointer exception on the first iteration of the while loop. (if (and (= (get-color (get-left @w)) black) (= (get-color (get-right @w)) black)) ;; Bug here! All of the code

Iterative Algorithm for Red-Black Tree

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问题 Can anyone please suggest me any pointer to an iterative algorithm for insertion and deletion into a Red-Black Tree? All the algorithms available in .Net/C# are based on recursion, which I can't trust for handling very large number of data (hence large number of recursion depth for insertion/deletion). Does anybody have one based on iteration? Note : Goletas.Collection uses an iterative algorithm for AVL tree which is highly efficient for large number of data, I want similar thing for Red

Can a red node have just 1 black child in a red-black tree?

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问题 The rules for a Red-Black Tree: Every node is either red or black. The root is black. Every leaf (NIL) is black. If a node is red, then both its children are black. For each node, all simple paths from the node to descendant leaves contain the same number of black nodes. Rule 4 mentions that red nodes need both black childs but what if there is just one child to begin with? Is there an argument to prove or disprove this? 回答1: No,a red node cannot have one child,consider the following cases:-

How to tell whether a red-black tree can have X black nodes and Y red nodes or not

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问题 I have an exam next week in algorithms, and was given questions to prepare for it. One of these questions has me stumped though. "Can we draw a red-black tree with 7 black nodes and 10 red nodes? why?" It sounds like it could be answered quickly, but I can't get my mind around it. The CRLS gives us the maximum height of a RB tree with n internal nodes: 2*lg(n+1). I think the problem could be solved using this lemma alone, but am not sure. Any tips? 回答1: Since this is exam preparation, I don't

How to check the black-height of a node for all paths to its descendent leaves?

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问题 Given a red-black tree , I need to write an efficient algorithm that checks whether for each node, all paths from from the node to descendant leaves contain the same number of black nodes, i.e. the algorithm should return a boolean if the property is true or false otherwise. 回答1: It wiil return the black height of the RB-tree. If the height is 0, the tree is an invalid red black tree. int BlackHeight(NodePtr root) { if (root == NULL) return 1; int leftBlackHeight = BlackHeight(root->left); if

Time complexity of TreeMap<> operations: get() and subMap()

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问题 Based on this post, Time complexity of TreeMap operations- subMap, headMap, tailMap subMap() itself is O(1), and O(n) comes from iterating the sub map. So, why use get(key) then? We can use subMap(key, true, key, true) instead, which is O(1) and iterating this sub map is also O(1). Faster than get(key), which is O(log(n)). Something wrong here... 回答1: We can use subMap(key, true, key, true) instead, which is O(1) This is correct and iterating this sub map is also O(1). O(n) comes from the

Red Black Tree ~ 1 Child Deletes

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问题 Is it ever possible for a red parent node to have just ONE black child node? I have been playing around with the Red/Black Tree simulator online and I can't manage to get a case of this to happen. The reason behind asking this is I believe I have an unnecessary IF in my code... if (temp_node->color == BLACK && node->color == RED) { node->color = BLACK; global_violation = false; } Thanks for any Feedback!! 回答1: No, this isn't possible. Remember, that in a red/black tree, all paths from the

How to check the black-height of a node for all paths to its descendent leaves?

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Given a red-black tree , I need to write an efficient algorithm that checks whether for each node, all paths from from the node to descendant leaves contain the same number of black nodes, i.e. the algorithm should return a boolean if the property is true or false otherwise. It wiil return the black height of the RB-tree. If the height is 0, the tree is an invalid red black tree. int BlackHeight(NodePtr root) { if (root == NULL) return 1; int leftBlackHeight = BlackHeight(root->left); if (leftBlackHeight == 0) return leftBlackHeight; int rightBlackHeight = BlackHeight(root->right); if

Can a red node have just 1 black child in a red-black tree?

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The rules for a Red-Black Tree: Every node is either red or black. The root is black. Every leaf (NIL) is black. If a node is red, then both its children are black. For each node, all simple paths from the node to descendant leaves contain the same number of black nodes. Rule 4 mentions that red nodes need both black childs but what if there is just one child to begin with? Is there an argument to prove or disprove this? No,a red node cannot have one child,consider the following cases:- 1.If the single child it has is red...this cannot happen because no two consecutive nodes can be red. 2.If