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  • RBTree(红黑树)

    http://www.linuxidc.com/Linux/2017-01/139950.htm

    http://blog.csdn.net/spch2008/article/details/9338923

    红黑树


           红黑树是一棵二叉搜索树,它在每个节点上增加了一个存储位来表示节点的颜色,可以是Red或Black。通过对任何一条从根到叶子简单路径上的颜色来约束,红黑树保证最长路径不超过最短路径的两倍,因而近似于平衡。


    红黑树是满足下面红黑性质的二叉搜索树
    1. 每个节点,不是红色就是黑色的;
    2. 根节点是黑色的;
    3. 如果一个节点是红色的,则它的两个子节点是黑色的;(从每个叶子到根的所有路径上不能有两个连续的红色节点)
    4. 对每个节点,从该节点到其所有后代叶节点的简单路径上,均包含相同数目的黑色节点;
    5. 每个叶子节点都是黑色的(这里的叶子节点是指的空节点)


    思考:为什么满足上面的颜色约束性质,红黑树能保证最长路径不超过最短路径的两倍?

    如图:所能增加的红节点数最多和黑节点数目一样多,故红黑树能保证最长路径不超过最短路径的两倍。

    这些约束强制了红黑树的关键性质: 从根到叶子的最长的可能路径不多于最短的可能路径的两倍长。结果是这个树大致上是平衡的。因为操作比如插入、删除和查找某个值的最坏情况时间都要求与树的高度成比例,这个在高度上的理论上限允许红黑树在最坏情况下都是高效的,而不同于普通的二叉查找树。
    要知道为什么这些特性确保了这个结果,注意到性质4导致了路径不能有两个毗连的红色节点就足够了。最短的可能路径都是黑色节点,最长的可能路径有交替的红色和黑色节点。因为根据性质5所有最长的路径都有相同数目的黑色节点,这就表明了没有路径能多于任何其他路径的两倍长。
    一、判断是否是红黑树:

     //判断是否是红黑树
        bool isRBTree()
        {
            int BlackNodeNum = 0;
            int curBlackNodeNum = 0;
            Node* cur = _root;
            while (cur)
            {
                if (cur->_col == BLACK)
                {
                    BlackNodeNum++;
                }
                cur = cur->_left;
            }
            return _isRBTree(_root, BlackNodeNum, curBlackNodeNum);
        }
    	    bool _isRBTree(Node* root, int BlackNodeNum, int curBlackNodeNum)
        {
            if (root == NULL)
            {
                return true;
            }
    
            if (root->_col == BLACK)
            {
                curBlackNodeNum++;
            }
    
            if (BlackNodeNum == curBlackNodeNum)
            {
                if (root->_parent == NULL)
                {
                    return true;
                }
                else if (root->_col == RED && root->_col == root->_parent->_col)
                {
                    return false;
                }
                else
                {
                    return true;
                }
            }
            return _isRBTree(root->_left, BlackNodeNum, curBlackNodeNum) 
    		&& _isRBTree(root->_right, BlackNodeNum, curBlackNodeNum);
        }
    三、左单旋  右边过多


        //左单旋 右边过多  红黑树放弃了追求完全平衡,追求大致平衡
        void RotateL(Node* &root) 
        {
            Node* subR = root->_right;
            Node* subRL = subR->_left;
    
            root->_right = subRL;
            if (subRL)
            {
                subRL->_parent = root;
            }
    
            subR->_left = root;
            subR->_parent = root->_parent;
            root->_parent = subR;
            root = subR;
    
            if (root->_parent == NULL)
            {
                _root = root;
            }
            else if (root->_parent->_key  >  root->_key)
            {
                root->_parent->_left = root;
            }
            else if ( root->_parent->_key < root->_key )
            {
                root->_parent->_right = root;
            }
        }

    四、右单旋 左边过多


     //右单旋 左边过多  红黑树放弃了追求完全平衡,追求大致平衡
        void RotateR(Node*& root)
        {
            Node* subL = root->_left;
            Node* subLR = subL->_right;
    
            root->_left = subLR;
            if (subLR)
            {
                subLR->_parent = root;
            }
    
            subL->_right = root;
            subL->_parent = root->_parent;
            root->_parent = subL;
    
            root = subL;
    
            if (root->_parent == NULL)
            {
                _root = parent;
            }
            else if (root->_parent->_key  >  root->_key)
            {
                root->_parent->_left = root;
            }
            else if (root->_parent->_key  <  root->_key)
            {
                root->_parent->_right = root;
            }
        }

    五、插入的三种情况
           ps:cur为当前节点,p为父节点,g为祖父节点,u为叔叔节点
    1.第一种情况
        cur为红,p为红,g为黑,u存在且为红,则将p,u改为黑,g改为红,然后把g当成cur,继续向上调整。



    2.第二种情况
       cur为红,p为红,g为黑,u不存在/u为黑
       p为g的左孩子,cur为p的左孩子,则进行右单旋转;相反,p为g的右孩子,cur为p的右孩子,则进行左单旋转,p、g变色--p变黑,g变红


    3.第三种情况
       cur为红,p为红,g为黑,u不存在/u为黑
       p为g的左孩子,cur为p的右孩子,则针对p做左单旋转;相反,p为g的右孩子,cur为p的左孩子,则针对p做右单旋转,则转换成了情况2


        上面已经把每种情况基本列出来了,其他相反的情况类似,反过来写一下就行了,具体详细过程参考代码。

        //红黑树的插入操作
        bool Insert(const K& key, const V& value)
        {
            if (_root == NULL)
            {
                _root = new Node(key, value);
                _root->_col = BLACK;
                return true;
            }
    
            Node* parent = NULL;
            Node* cur = _root;
            while (cur)
            {
                if (cur->_key > key)
                {
                    parent = cur;
                    cur = cur->_left;
                }
                else if (cur->_key < key)
                {
                    parent = cur;
                    cur = cur->_right;
                }
                else
                {
                    return false;
                }
            }
    
            //插入位置
            if (parent->_key >key)
            {
                cur = new Node(key, value);
                parent->_left = cur;
                cur->_parent = parent;
            }
            else if (parent->_key < key)
            {
                cur = new Node(key, value);
                parent->_right = cur;
                cur->_parent = parent;
            }
    
            //插入以后,进行调整
            while (cur != _root && parent->_col == RED)
            {
                Node* grandfather = parent->_parent;
                Node* uncle = NULL;
                //左边的情况
                if (parent == grandfather->_left)
                {
                    //情况一
                    uncle = grandfather->_right;
                    if (uncle && uncle->_col == RED)
                    {
                        //1. 不需要旋转
                        if (cur == parent->_left)
                        {
                            grandfather->_col = RED;
                            parent->_col = BLACK;
                            uncle->_col = BLACK;
    
                            cur = grandfather;
                            parent = cur->_parent;
                        }
    
                        //2.需要旋转
                        else if (cur == parent->_right)
                        {
                            RotateL(parent);
                            grandfather->_col = RED;
                            parent->_col = BLACK;
                            uncle->_col = BLACK;
    
                            cur = grandfather;
                            parent = cur->_parent;
                        }
    
                    }
    
                    //情况二,三
                    else if (uncle == NULL || (uncle && uncle->_col == BLACK))
                    {
                        if (cur == parent->_right)
                        {
                            RotateL(parent);
                        }
                        parent->_col = BLACK;
                        grandfather->_col = RED;
                        RotateR(grandfather);
                        break;
                    }
                }
                //右边的情况
                else if (parent == grandfather->_right)
                {
                    //情况一
                    uncle = grandfather->_left;
                    if (uncle && uncle->_col == RED)
                    {
                        //1.不需要旋转
                        if (cur == parent->_right)
                        {
                            uncle->_col = BLACK;
                            grandfather->_col = RED;
                            parent->_col = BLACK;
    
                            cur = grandfather;
                            parent = cur->_parent;
                        }
    
                        //2.需要旋转
                        else if (cur == parent->_left)
                        {
    						RotateR(parent);
                            uncle->_col = BLACK;
                            grandfather->_col = RED;
                            parent->_col = BLACK;
    
                            cur = grandfather;
                            parent = cur->_parent;
                        }
                    }
                    //情况二,三
                    else if (uncle == NULL || (uncle && uncle->_col == BLACK))
                    {
                        if (cur == parent->_left)
                        {
                            RotateR(parent);
                        }
                        parent->_col = BLACK;
                        grandfather->_col = RED;
                        RotateL(grandfather);
                        break;
                    }
                }
            }
            _root->_col = BLACK;
            return true;
        }
    
            红黑树和AVL树都是高效的平衡二叉树,增删查改的时间复杂度都是O(lg(N)),红黑树的不追求完全平衡,保证最长路径不超过最短路径的2倍,相对而言,降低了旋转的要求,所以性能会优于AVL树,所以实际运用中红黑树更多。

    #include<iostream>
    using namespace std;
    
    enum colour
    {
        RED,
        BLACK,
    };
    
    template<class K, class V>
    struct RBTreeNode
    {
        int _col;
        K _key;
        V _value;
        RBTreeNode<K, V>* _left;
        RBTreeNode<K, V>* _right;
        RBTreeNode<K, V>* _parent;
    
        RBTreeNode(const K& key, const V& value) :
            _key(key),  _value(value), _col(RED), _left(NULL), _right(NULL), _parent(NULL)
        {
    
        }
    };
    
    template<class K, class V>
    class RBTree
    {
        typedef  RBTreeNode<K, V>  Node;
    public:
        RBTree():_root(NULL)
        {
    
        }
        //红黑树的插入操作
        bool Insert(const K& key, const V& value)
        {
            if (_root == NULL)
            {
                _root = new Node(key, value);
                _root->_col = BLACK;
                return true;
            }
    
            Node* parent = NULL;
            Node* cur = _root;
            while (cur)
            {
                if (cur->_key > key)
                {
                    parent = cur;
                    cur = cur->_left;
                }
                else if (cur->_key < key)
                {
                    parent = cur;
                    cur = cur->_right;
                }
                else
                {
                    return false;
                }
            }
    
            //插入位置
            if (parent->_key >key)
            {
                cur = new Node(key, value);
                parent->_left = cur;
                cur->_parent = parent;
            }
            else if (parent->_key < key)
            {
                cur = new Node(key, value);
                parent->_right = cur;
                cur->_parent = parent;
            }
    
            //插入以后,进行调整
            while (cur != _root && parent->_col == RED)
            {
                Node* grandfather = parent->_parent;
                Node* uncle = NULL;
                //左边的情况
                if (parent == grandfather->_left)
                {
                    //情况一
                    uncle = grandfather->_right;
                    if (uncle && uncle->_col == RED)
                    {
                        //1. 不需要旋转
                        if (cur == parent->_left)
                        {
                            grandfather->_col = RED;
                            parent->_col = BLACK;
                            uncle->_col = BLACK;
    
                            cur = grandfather;
                            parent = cur->_parent;
                        }
    
                        //2.需要旋转
                        else if (cur == parent->_right)
                        {
                            RotateL(parent);
                            grandfather->_col = RED;
                            parent->_col = BLACK;
                            uncle->_col = BLACK;
    
                            cur = grandfather;
                            parent = cur->_parent;
                        }
    
                    }
    
                    //情况二,三
                    else if (uncle == NULL || (uncle && uncle->_col == BLACK))
                    {
                        if (cur == parent->_right)
                        {
                            RotateL(parent);
                        }
                        parent->_col = BLACK;
                        grandfather->_col = RED;
                        RotateR(grandfather);
                        break;
                    }
                }
                //右边的情况
                else if (parent == grandfather->_right)
                {
                    //情况一
                    uncle = grandfather->_left;
                    if (uncle && uncle->_col == RED)
                    {
                        //1.不需要旋转
                        if (cur == parent->_right)
                        {
                            uncle->_col = BLACK;
                            grandfather->_col = RED;
                            parent->_col = BLACK;
    
                            cur = grandfather;
                            parent = cur->_parent;
                        }
    
                        //2.需要旋转
                        else if (cur == parent->_left)
                        {
                            RotateR(parent);
                            uncle->_col = BLACK;
                            grandfather->_col = RED;
                            parent->_col = BLACK;
    
                            cur = grandfather;
                            parent = cur->_parent;
                        }
                    }
                    //情况二,三
                    else if (uncle == NULL || (uncle && uncle->_col == BLACK))
                    {
                        if (cur == parent->_left)
                        {
                            RotateR(parent);
                        }
                        parent->_col = BLACK;
                        grandfather->_col = RED;
                        RotateL(grandfather);
                        break;
                    }
                }
            }
            _root->_col = BLACK;
            return true;
        }
    
        //判断是否是红黑树
        bool isRBTree()
        {
            int BlackNodeNum = 0;
            int curBlackNodeNum = 0;
            Node* cur = _root;
            while (cur)
            {
                if (cur->_col == BLACK)
                {
                    BlackNodeNum++;
                }
                cur = cur->_left;
            }
            return _isRBTree(_root, BlackNodeNum, curBlackNodeNum);
        }
    
        //中序遍历
        void InOrderTraverse()
        {
            _InOrder(_root);
        }
    
    protected:
        bool _isRBTree(Node* root, int BlackNodeNum, int curBlackNodeNum)
        {
            if (root == NULL)
            {
                return true;
            }
    
            if (root->_col == BLACK)
            {
                curBlackNodeNum++;
            }
    
            if (BlackNodeNum == curBlackNodeNum)
            {
                if (root->_parent == NULL)
                {
                    return true;
                }
                else if (root->_col == RED && root->_col == root->_parent->_col)
                {
                    return false;
                }
                else
                {
                    return true;
                }
            }
            return _isRBTree(root->_left, BlackNodeNum, curBlackNodeNum) && _isRBTree(root->_right, BlackNodeNum, curBlackNodeNum);
        }
    
    
        void _InOrder(Node* root)
        {
            if (root)
            {
                _InOrder(root->_left);
                cout << root->_key << " ";
                _InOrder(root->_right);
            }
        }
    
        //左单旋 右边过多  红黑树放弃了追求完全平衡,追求大致平衡
        void RotateL(Node* &root)
        {
            Node* subR = root->_right;
            Node* subRL = subR->_left;
    
            root->_right = subRL;
            if (subRL)
            {
                subRL->_parent = root;
            }
    
            subR->_left = root;
            subR->_parent = root->_parent;
            root->_parent = subR;
            root = subR;
    
            if (root->_parent == NULL)
            {
                _root = root;
            }
            else if (root->_parent->_key  >  root->_key)
            {
                root->_parent->_left = root;
            }
            else if ( root->_parent->_key < root->_key )
            {
                root->_parent->_right = root;
            }
        }
    
        //右单旋 左边过多  红黑树放弃了追求完全平衡,追求大致平衡
        void RotateR(Node*& root)
        {
            Node* subL = root->_left;
            Node* subLR = subL->_right;
    
            root->_left = subLR;
            if (subLR)
            {
                subLR->_parent = root;
            }
    
            subL->_right = root;
            subL->_parent = root->_parent;
            root->_parent = subL;
    
            root = subL;
    
            if (root->_parent == NULL)
            {
                _root = parent;
            }
            else if (root->_parent->_key  >  root->_key)
            {
                root->_parent->_left = root;
            }
            else if (root->_parent->_key  <  root->_key)
            {
                root->_parent->_right = root;
            }
        }
    protected:
        Node* _root;
    };
    

    #include "Queue.h"
    void TestRBtree()
    {
        RBTree<int, int>RBT;
        int Arr[10]= { 1, 2, 5, 12, 16, 18, 26, 3, 99, 100 };
    
        for (int i = 0; i < 10; i++)
        {
            RBT.Insert(Arr[i], i);
        }
        RBT.InOrderTraverse();
        cout << endl;
        cout << "is RBTree ?: " << RBT.isRBTree() << endl;
    
    }
    
    int main()
    {
        TestRBtree();
        return 0;
    }
    



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  • 原文地址:https://www.cnblogs.com/sowhat1412/p/12734447.html
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