随机构建的二叉查找树的高度期望值为O(lgn),并不代表所有的二叉查找树的高度都为O(lgn)。但是对于有些二叉查找树的变形来说,动态集合各基本操作的性能却总是很好的,如红黑树、B树、平衡二叉树(AVL树)、跳跃表(确切的说不是树,或多或少有树的结构)、treaps(树堆)等。这里我们讲解红黑树。
平衡的意思就是完成动态数据集的操作(minimum、maximum、search、predecessor(前驱)、successor(后继)、insert及delete)在O(lgn)时间内完成。
定义:
红黑树在二叉查找树的基础上增加了色域-red或者black。 一棵红黑树要满足下面4条性质:(1)结点要么是红的,要么是黑的。
(2)根和叶子节点都是黑的。
(3)一个结点是红的,那么它的父节点必须是黑的。
(4)任一结点到其叶子节点的简单路径上黑结点的个数要相同= 黑高度(不包括自身结点,但包括叶子节点)。
性质:
一个具有n个结点(不包括叶子节点)的红黑树的高度不超过2lg(n+1)。
因此这些动态集合的操作都在O(lgn)时间内能完成,所以红黑树是平衡的。
操作(插入和删除)
红黑树操作与二叉查找树的主要区别在于插入和删除,其它都是一样的。可参见上一篇blog。
插入:
并且红黑树只能通过插入操作来创建一棵红黑树以维持红黑特性。红黑树的插入操作分为三种情况.如下图
待传…..
其中case1为着色,可以一直可以向上移动,每次移动2层,由于树高≤2lg(n+1),所以最多循环lg(n+1)次。case2和case3是终止情形,各旋转一次,都为O(1)时间。所以红黑树的插入操作可以在O(lgn)时间内完成.
代码如下:
RBTree.h
#ifndef RBTREE #define RBTREE #include <iostream> #include <string> using namespace std; class BSTNode{ public: BSTNode *left, *right; BSTNode *parent; int val; string color; }; class RBTree{ public: RBTree(const int &rootVal){ root = new BSTNode(); root->val = rootVal; root->parent = NULL; root->left = NULL; root->right = NULL; root->color = "black"; } void insertRBTree(BSTNode *root, const int &value); // 红黑树的插入 void inorderRBTree(BSTNode *p); BSTNode *root; private: BSTNode *insertBST(BSTNode *p, const int &value); // 二叉树的插入 }; #endif
#include "RBTree.h" using namespace std; BSTNode* RBTree::insertBST(BSTNode *p, const int &value){ BSTNode *y = NULL; BSTNode *in = new BSTNode(); in->left = NULL; in->right = NULL; in->val = value; in->parent = NULL; while(p != NULL){ y = p; if(p->val > in->val) p = p->left; else p = p->right; } if(y == NULL) p = in; else{ in->parent = y; if(y->val > in->val) y->left = in; else y->right = in; } return in; } void RBTree::insertRBTree(BSTNode *root1, const int &value){ BSTNode * in = insertBST(root1, value); in->color = "red"; while(in != root1 && in->color == "red"){ // 对红黑特性进行调整 if(in->parent->color == "black") return; // 也就保证了必须 if(in->parent == in->parent->parent->left){ BSTNode *y = in->parent->parent->right; if(y != NULL && y->color == "red"){ // case 1 y->color = "black"; y->parent->color = "red"; in->parent->color ="black"; in = in->parent->parent; } else{ if(in == in->parent->right){ // case 2 in->parent 左旋 BSTNode *pa = in->parent; in->parent = pa->parent; pa->parent->left = in; pa->parent = in; if(in->left != NULL) in->left->parent = pa; pa->right = in->left; in->left = pa; in = pa; } // case 3 in->parent->parent 右旋 BSTNode *pa = in->parent; BSTNode *gpa = in->parent->parent; if(gpa->parent != NULL){ if(gpa == gpa->parent->left){ gpa->parent->left = pa; }else gpa->parent->right = pa; } pa->parent = gpa->parent; if(pa->right != NULL) pa->right->parent = gpa; gpa->left = pa->right; pa->right = gpa; gpa->parent = pa; pa->color = "black"; gpa->color = "red"; } } else{ BSTNode *y = in->parent->parent->left; if(y != NULL && y->color == "red"){ // case 1 y->color = "black"; y->parent->color = "red"; in->parent->color ="black"; in = in->parent->parent; }else{ // do the same as A but left与right对换 if(in == in->parent->left){ // case 2 in->parent 右旋 BSTNode *pa = in->parent; in->parent = pa->parent; pa->parent->right = in; pa->parent = in; if(in->right != NULL) in->right->parent = pa; pa->left = in->right; in->right = pa; in = pa; } // case 3 in->parent->parent 左旋 BSTNode *pa = in->parent; BSTNode *gpa = in->parent->parent; if(gpa->parent != NULL){ if(gpa == gpa->parent->left){ gpa->parent->left = pa; }else gpa->parent->right = pa; } pa->parent = gpa->parent; if(pa->left != NULL) pa->left->parent = gpa; gpa->right = pa->left; pa->left = gpa; gpa->parent = pa; pa->color = "black"; gpa->color = "red"; } } } root1->color = "black"; } void RBTree::inorderRBTree(BSTNode *p){ if(p == NULL) return; if(p->left != NULL) inorderRBTree(p->left); cout << p->val << p->color << " "; if(p->right != NULL) inorderRBTree(p->right); }
Main.cpp
int a[10] = {5,4,6, 7,2,4, 1, 8, 5, 10}; RBTree rbt(a[0]); for(int i =1; i < 10; i++) rbt.insertRBTree(rbt.root, a[i]); cout << "中序遍历的结果: " << endl; rbt.inorderRBTree(rbt.root);Result:
删除:
待续…
原文地址:http://blog.csdn.net/lu597203933/article/details/43458037