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这篇是红黑树整体的模拟实现和讲解
一：红黑树的封装与适配
- 代码：
//颜色
enum Colour
{
RED,
BLACK,
};
template<class T>
struct RBTreeNode
{
RBTreeNode<T>* _left;
RBTreeNode<T>* _right;
RBTreeNode<T>* _parent;
T _data;//T可以是key也可以是pair<K,V>
Colour _col;
RBTreeNode(const T& data)
:_left(nullptr)
, _right(nullptr)
, _parent(nullptr)
, _data(data)
, _col(RED)
{}
};
template<class K, class T, class KeyOfT>
class RBTree
{
typedef RBTreeNode<T> Node;
public:
typedef _TreeIterator<T, T&, T*> iterator;
typedef _TreeIterator<T,const T&, const T*> const_iterator;
typedef ReverseIterator<iterator> reverse_iterator;
typedef ReverseIterator<const_iterator> reverse_const_iterator;
RBTree()
:_root(nullptr)
{}
~RBTree()
{
_Destory(_root);
}
iterator begin()
{
Node* cur = _root;
while (cur&&cur->_left)
{
cur = cur->_left;
}
return iterator(cur);
}
reverse_iterator rbegin()
{
Node* cur = _root;
while (cur&&cur->_right)
{
cur = cur->_right;
}
return reverse_iterator(iterator(cur));
}
reverse_iterator rend()
{
return reverse_iterator(iterator(nullptr));
}
iterator end()
{
return iterator(nullptr);
}
Node* Find(const K& key)
{
KeyOfT kot;
Node* cur = _root;
while (cur)
{
if (kot(cur->_kv.first) > key)
{
cur = cur->_left;
}
else if (kot(cur->_kv.first) < key)
{
cur = cur->_right;
}
else
{
return cur;
}
}
return nullptr;
}
pair<iterator, bool> Insert(const T& data)
{
//空树的情况
if (_root == nullptr)
{
_root = new Node(data);
_root->_col = BLACK;
return make_pair(iterator(_root), true);
}
KeyOfT kot;
//查找位置插入节点
Node* cur = _root, * parent = _root;
while (cur)
{
if (kot(cur->_data) > kot(data))
{
parent = cur;
cur = cur->_left;
}
else if (kot(cur->_data) < kot(data))
{
parent = cur;
cur = cur->_right;
}
else
{
return make_pair(iterator(cur), false);
}
}
//创建链接节点
cur = new Node(data);
Node* newnode = cur;
if (kot(parent->_data) > kot(data))
{
parent->_left = cur;
}
else
{
parent->_right = cur;
}
cur->_parent = parent;
//父节点存在且为红，则需要调整（不能存在连续的红色节点）
while (parent && parent->_col == RED)
{
//此时当前节点一定有祖父节点
Node* granparent = parent->_parent;
//具体调整情况主要看叔叔节点
//分左右讨论
if (parent == granparent->_left)
{
Node* uncle = granparent->_right;
//情况1：叔叔节点存在且为红
if (uncle && uncle->_col == RED)
{
//修改颜色，继续向上检查
granparent->_col = RED;
parent->_col = uncle->_col = BLACK;
cur = granparent;
parent = cur->_parent;
}
else//情况2和3：叔叔节点不存在 或者存在且为黑
{
//单旋(三代节点为斜线) 变色
if (cur == parent->_left)
{
RotateR(granparent);
granparent->_col = RED;
parent->_col = BLACK;
}
else//双旋(三代节点为折线) 变色
{
RotateL(parent);
RotateR(granparent);
cur->_col = BLACK;
granparent->_col = RED;
}
//旋转后不需再向上调整了
break;
}
}
else//parent=grandparent->right
{
Node* uncle = granparent->_left;
if (uncle && uncle->_col == RED)
{
parent->_col = uncle->_col = BLACK;
granparent->_col = RED;
cur = granparent;
parent = cur->_parent;
}
else
{
if (cur == parent->_right)
{
RotateL(granparent);
parent->_col = BLACK;
granparent->_col = RED;
}
else
{
RotateR(parent);
RotateL(granparent);
cur->_col = BLACK;
granparent->_col = RED;
}
break;
}
}
}
//确保根节点为黑
_root->_col = BLACK;
return make_pair(iterator(newnode), true);
}
bool IsRBTree()
{
if (_root == nullptr)
{
return true;
}
if (_root->_col == RED)
{
cout << "根节点为红色" << endl;
return false;
}
int Blacknum = 0;
Node* cur = _root;
while (cur)
{
if (cur->_col == BLACK)
Blacknum ;
cur = cur->_left;
}
int i = 0;
return _IsRBTree(_root, Blacknum, i);
}
private:
void _Destory(Node*& root)
{
if (root == nullptr)
return;
_Destory(root->_left);
_Destory(root->_right);
delete root;
root = nullptr;
}
bool _IsRBTree(Node* root, int blacknum, int count)
{
if (root == nullptr)
{
if (blacknum == count)
return true;
cout << "各路径上黑色节点个数不同" << endl;
return false;
}
if (root->_col == RED && root->_parent->_col == RED)
{
cout << "存在连续红色节点" << endl;
return false;
}
if (root->_col == BLACK)
count ;
return _IsRBTree(root->_left, blacknum, count) && _IsRBTree(root->_right, blacknum, count);
}
void RotateL(Node* parent)
{
Node* subR = parent->_right;
Node* subRL = subR->_left;
Node* parentP = parent->_parent;
parent->_right = subRL;
if (subRL)
{
subRL->_parent = parent;
}
subR->_left = parent;
parent->_parent = subR;
if (parent == _root)
{
_root = subR;
subR->_parent = nullptr;
}
else
{
subR->_parent = parentP;
if (parentP->_left == parent)
{
parentP->_left = subR;
}
else
{
parentP->_right = subR;
}
}
}
void RotateR(Node* parent)
{
Node* subL = parent->_left;
Node* subLR = subL->_right;
Node* parentP = parent->_parent;
parent->_left = subLR;
if (subLR)
{
subLR->_parent = parent;
}
subL->_right = parent;
parent->_parent = subL;
if (parent == _root)
{
_root = subL;
subL->_parent = nullptr;
}
else
{
subL->_parent = parentP;
if (parentP->_left == parent)
{
parentP->_left = subL;
}
else
{
parentP->_right = subL;
}
}
}
private:
Node* _root;
};二：map的封装和模拟实现
- 代码：
namespace ymhh
{
template<class K, class V>
class map
{
struct MapOfKey
{
const K& operator()(const pair<K, V>& kv)
{
return kv.first;
}
};
public:
typedef typename RBTree<K, pair<const K, V>, MapOfKey>::iterator iterator;
typedef typename RBTree<K, pair<const K, V>, MapOfKey>::reverse_iterator reverse_iterator;
iterator begin()
{
return _t.begin();
}
iterator end()
{
return _t.end();
}
reverse_iterator rbegin()
{
return _t.rbegin();
}
reverse_iterator rend()
{
return _t.rend();
}
pair<iterator, bool> insert(const pair<const K, V>& kv)
{
return _t.Insert(kv);
}
V& operator[](const K& key)
{
pair<iterator, bool> ret = insert(make_pair(key, V()));
return ret.first->second;
}
iterator find(const K& key)
{
return _t.Find(key);
}
private:
RBTree<K, pair<const K, V>, MapOfKey> _t;
};
}三： set的封装和模拟实现
- 代码：
namespace ymhh
{
template<class K>
class set
{
struct SetOfKey
{
const K& operator()(const K& key)
{
return key;
}
};
public:
typedef typename RBTree<K,K, SetOfKey>::iterator iterator;
typedef typename RBTree<K,K, SetOfKey>::reverse_iterator reverse_iterator;
iterator begin()
{
return _t.begin();
}
iterator end()
{
return _t.end();
}
reverse_iterator rbegin()
{
return _t.rbegin();
}
reverse_iterator rend()
{
return _t.rend();
}
pair<iterator, bool> insert(const K& key)
{
return _t.Insert(key);
}
iterator find(const K& key)
{
return _t.Find(key);
}
private:
RBTree<K, K, SetOfKey> _t;
};
}总结
因为红黑树的增删查改都是O（logN）,所以用红黑树实现的map/set的增删查改也是O（logN），是个非常优秀的容器
