目录
红黑树的概念
红黑树的性质
红黑树节点的定义
红黑树的插入
1. 按照二叉搜索的树规则插入新节点
2. 检测新节点插入后,红黑树的性质是否造到破坏
红黑树的检测
红黑树的删除
红黑树和AVL树的比较
红黑树的概念
红黑树,是一种二叉搜索树,但在每个结点上增加一个存储位表示结点的颜色,可以是Red或 Black。 通过对任何一条从根到叶子的路径上各个结点着色方式的限制,红黑树确保没有一条路 径会比其他路径长出俩倍,因而是接近平衡的。
红黑树的性质
1. 每个结点不是红色就是黑色
2. 根节点是黑色的
3. 如果一个节点是红色的,则它的两个孩子结点是黑色的
4. 对于每个结点,从该结点到其所有后代叶结点的简单路径上,均 包含相同数目的黑色结点
5. 每个叶子结点都是黑色的(此处的叶子结点指的是空结点)
思考:为什么满足上面的性质,红黑树就能保证:其最长路径中节点个数不会超过最短路径节点 个数的两倍?
红黑树节点的定义
//节点的颜色
enum Colour
{BLACK, RED
};//节点定义
template<class T>
struct RBTreenode
{ T _data; //储存的数据RBTreenode<T>* _left; //左孩子RBTreenode<T>* _right; //右孩子RBTreenode<T>* _parent; //父亲Colour _col; //颜色//构造函数RBTreenode(const T& data):_data(data), _left(nullptr), _right(nullptr), _parent(nullptr), _col(RED){}};
再我们定义节点的时候,需要用一个枚举类型来定义我们的节点的颜色。然后还是三叉链,左孩子,右孩子,父亲。
红黑树的插入
红黑树是在二叉搜索树的基础上加上其平衡限制条件,因此红黑树的插入可分为两步:
1. 按照二叉搜索的树规则插入新节点
bool Insert(const pair<K, V>& kv)
{ //如果跟为空,则直接当做根if (_root == nullptr){_root = new node(kv);_root->_col = BLACK;return true;} //根不为空,按二叉搜索树的规则插入node* parent = nullptr;node* cur = _root;while (cur){if (kv.first > cur->_kv.first){parent = cur;cur = cur->_right;}else if(kv.first<cur->_kv.first){parent = cur;cur = cur->_left;}else{return false;}}cur = new node(kv);if (kv.first < parent->_kv.first)parent->_left = cur;elseparent->_right = cur;cur->_parent = parent;// ~~~~~~~~~~~~~~~~~~~~~~//判断颜色return true;
}
2. 检测新节点插入后,红黑树的性质是否造到破坏
因为新节点的默认颜色是红色,因此:如果其双亲节点的颜色是黑色,没有违反红黑树任何 性质,则不需要调整;但当新插入节点的双亲节点颜色为红色时,就违反了性质三不能有连 在一起的红色节点,此时需要对红黑树分情况来讨论:
约定:cur为当前节点,p为父节点,g为祖父节点,u为叔叔节点
情况一: cur为红,p为红,g为黑,u存在且为红
看到上图,cur插入的时候本身就为红,如果说它的父亲为红,叔叔存在且为红的话,我们就需要把p和u 变为黑,再让g 变为红,然后我们还需要向上更新,因为这幅图可能是一颗子树,它的祖先也需要更新颜色,如下图。
情况二: cur为红,p为红,g为黑,u不存在/u存在且为黑
说明: u的情况有两种
1.如果u节点存在,则cur一定是新插入的节点,因为如果cur不是新插入的节点,则cur和p一定有一个的节点是黑色,就不满足性质4:每条路径黑色节点个数相同。
2. 如果u节点存在,则其一定是黑色的,那么cur节点原来的颜色一定是黑色的,现在看到其是红色的原因是因为cur的子树在调整的过程中将cur由黑色变为了红色。
对于这种情况,我们就需要进行旋转了,对,没错,和AVL树里面的旋转操作是一样的,代码我就复制一下。
p为g的左孩子,cur为p的左孩子,则进行右单旋;
相反,p为g的右孩子,cur为p的右孩子,则进行左单旋。
我们在旋转完成之后,需要和上图一样,再把相应的颜色变换就行了。
右单旋:
void RotateR(node* parent)
{node* subL = parent->_left;node* subLR = subL->_right;parent->_left = subLR;if (subLR)subLR->_parent = parent;node* pParent = parent->_parent;subL->_right = parent;parent->_parent = subL;if (parent == _root){_root = subL;subL->_parent = nullptr;}else{if (pParent->_left == parent){pParent->_left = subL;}else{pParent->_right = subL;}subL->_parent = pParent;}
}
左单旋:
void RotateL(node* parent)
{node* subR = parent->_right;node* subRL = subR->_left;parent->_right = subRL;if (subRL)subRL->_parent = parent;node* parentParent = parent->_parent;subR->_left = parent;parent->_parent = subR;if (parentParent == nullptr){_root = subR;subR->_parent = nullptr;}else{if (parent == parentParent->_left){parentParent->_left = subR;}else{parentParent->_right = subR;}subR->_parent = parentParent;}
}
情况三: cur为红,p为红,g为黑,u不存在/u存在且为黑
像上面这种情况,我们就需要旋转两次,和AVL树的操作差不多。
p为g的左孩子,cur为p的右孩子,则对p进行左单旋;
相反,p为g的右孩子,cur为p的左孩子,则对p进行右单旋。
注意:这样它们就转化为了情况二,所以,我们再对它们进行情况二的右单旋或左单旋即可。旋转完之后,再修改它们的颜色即可。
我们在插入的时候,对每种情况进行相应的处理即可。
bool Insert(const pair<K, V>& kv)
{if (_root == nullptr){_root = new node(kv);_root->_col = BLACK;return true;}node* parent = nullptr;node* cur = _root;while (cur){if (kv.first > cur->_kv.first){parent = cur;cur = cur->_right;}else if(kv.first<cur->_kv.first){parent = cur;cur = cur->_left;}else{return false;}}cur = new node(kv);if (kv.first < parent->_kv.first)parent->_left = cur;elseparent->_right = cur;cur->_parent = parent;//判断颜色 ,变色while (parent&&parent->_col == RED){node* grandfather = parent->_parent;if (parent == grandfather->_left){// g//p unode* uncle = grandfather->_right;//叔叔存在且为红,只需变色,再向上更新即可if (uncle && uncle->_col == RED){parent->_col = uncle->_col = BLACK;grandfather->_col = RED;//向上更新cur = grandfather;parent = cur->_parent;}//叔叔不存在,或者存在且为黑(旋转再变色)else{if (cur == parent->_left){// g// p u// cRotateR(grandfather);grandfather->_col = RED;parent->_col = BLACK;}else{// g// p u// cRotateL(parent);RotateR(grandfather);grandfather->_col = RED;cur->_col = BLACK;}break;}}else //parent == grandfather->_right{// g// u pnode* uncle = grandfather->_left;if (uncle && uncle->_col == RED){parent->_col = uncle->_col = BLACK;grandfather->_col = RED;//向上更新cur = grandfather;parent = cur->_parent;}else{if (cur == parent->_right){RotateL(grandfather);grandfather->_col = RED;parent->_col = BLACK;}else{RotateR(parent);RotateL(grandfather);cur->_col = BLACK;grandfather->_col = RED;}break;}}_root->_col = BLACK;}return true;
}
以上就是整个的插入代码。
红黑树的检测
红黑树的检测分为两步:
1. 检测其是否满足二叉搜索树(中序遍历是否为有序序列)
2. 检测其是否满足红黑树的性质
中序遍历
void _Inorder(node* _root){if (_root == nullptr){return;}_Inorder(_root->_left);cout << _root->_kv.first << ":" << _root->_kv.second << endl;_Inorder(_root->_right);}
检查
bool Check(node* _root, int blackNum, const int refNum)
{if (_root == nullptr){if (blackNum != refNum){cout << "存在路径黑色节点不同" << endl;return false;}return true;}if (_root->_col == BLACK)blackNum++;if (_root->_col == RED && _root->_parent->_col == RED){cout << "存在两个连续节点都为红色" << endl;return false;}return Check(_root->_left, blackNum, refNum) && Check(_root->_right, blackNum, refNum);
}
红黑树的删除
https://www.cnblogs.com/fornever/archive/2011/12/02/2270692.html
删除操作我们只做了解就行了。
红黑树和AVL树的比较
红黑树和AVL树都是高效的平衡二叉树,增删改查的时间复杂度都是O($log_2 N$),红黑树不追 求绝对平衡,其只需保证最长路径不超过最短路径的2倍,相对而言,降低了插入和旋转的次数, 所以在经常进行增删的结构中性能比AVL树更优,而且红黑树实现比较简单,所以实际运用中红 黑树更多。
总代码
enum Colour
{BLACK,RED
};template<class K,class V>
struct RBTreenode
{pair < K, V> _kv;RBTreenode<K, V>* _left;RBTreenode<K, V>* _right;RBTreenode<K, V>* _parent;Colour _col;RBTreenode(const pair<K, V>& kv):_kv(kv), _left(nullptr), _right(nullptr), _parent(nullptr), _col(RED){}};template<class K,class V>
class RBTree
{typedef RBTreenode<K, V> node;public:bool Insert(const pair<K, V>& kv){if (_root == nullptr){_root = new node(kv);_root->_col = BLACK;return true;}node* parent = nullptr;node* cur = _root;while (cur){if (kv.first > cur->_kv.first){parent = cur;cur = cur->_right;}else if(kv.first<cur->_kv.first){parent = cur;cur = cur->_left;}else{return false;}}cur = new node(kv);if (kv.first < parent->_kv.first)parent->_left = cur;elseparent->_right = cur;cur->_parent = parent;//判断颜色 ,变色while (parent&&parent->_col == RED){node* grandfather = parent->_parent;if (parent == grandfather->_left){// g//p unode* uncle = grandfather->_right;//叔叔存在且为红,只需变色,再向上更新即可if (uncle && uncle->_col == RED){parent->_col = uncle->_col = BLACK;grandfather->_col = RED;//向上更新cur = grandfather;parent = cur->_parent;}//叔叔不存在,或者存在且为黑(旋转再变色)else{if (cur == parent->_left){// g// p u// cRotateR(grandfather);grandfather->_col = RED;parent->_col = BLACK;}else{// g// p u// cRotateL(parent);RotateR(grandfather);grandfather->_col = RED;cur->_col = BLACK;}break;}}else //parent == grandfather->_right{// g// u pnode* uncle = grandfather->_left;if (uncle && uncle->_col == RED){parent->_col = uncle->_col = BLACK;grandfather->_col = RED;//向上更新cur = grandfather;parent = cur->_parent;}else{if (cur == parent->_right){RotateL(grandfather);grandfather->_col = RED;parent->_col = BLACK;}else{RotateR(parent);RotateL(grandfather);cur->_col = BLACK;grandfather->_col = RED;}break;}}_root->_col = BLACK;}return true;}void RotateR(node* parent){node* subL = parent->_left;node* subLR = subL->_right;parent->_left = subLR;if (subLR)subLR->_parent = parent;node* pParent = parent->_parent;subL->_right = parent;parent->_parent = subL;if (parent == _root){_root = subL;subL->_parent = nullptr;}else{if (pParent->_left == parent){pParent->_left = subL;}else{pParent->_right = subL;}subL->_parent = pParent;}}void RotateL(node* parent){node* subR = parent->_right;node* subRL = subR->_left;parent->_right = subRL;if (subRL)subRL->_parent = parent;node* parentParent = parent->_parent;subR->_left = parent;parent->_parent = subR;if (parentParent == nullptr){_root = subR;subR->_parent = nullptr;}else{if (parent == parentParent->_left){parentParent->_left = subR;}else{parentParent->_right = subR;}subR->_parent = parentParent;}}void Inorder(){_Inorder(_root);cout << endl;}int size(){return _size(_root);}int height(){return _height(_root);}bool IsBalance(){if (_root == nullptr)return true;if (_root->_col == RED)return false;int refnum = 0;node* cur = _root;while (cur){if (cur->_col == BLACK)refnum++;cur = cur->_left;}return Check(_root, 0, refnum);}private:bool Check(node* _root, int blackNum, const int refNum){if (_root == nullptr){if (blackNum != refNum){cout << "存在路径黑色节点不同" << endl;return false;}return true;}if (_root->_col == BLACK)blackNum++;if (_root->_col == RED && _root->_parent->_col == RED){cout << "存在两个连续节点都为红色" << endl;return false;}return Check(_root->_left, blackNum, refNum) && Check(_root->_right, blackNum, refNum);}int _height(node* _root){if (_root == nullptr)return 0;int lefth = _height(_root->_left);int righth = _height(_root->_right);return lefth > righth ? lefth + 1 : righth + 1;}int _size(node*_root){if (_root == nullptr)return 0;return _size(_root->_left) + _size(_root->_right) + 1;}void _Inorder(node* _root){if (_root == nullptr){return;}_Inorder(_root->_left);cout << _root->_kv.first << ":" << _root->_kv.second << endl;_Inorder(_root->_right);}node* _root=nullptr;};
我们后面还会用红黑树来模拟实现map和set。