【数据结构】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(没有连续的红节点)
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。（每条路径的黑色节点的数量相等）

Test.cpp（主函数）

#include<iostream>
using namespace std;

#include"RBTree.h"

int main()
{
	TestTree();
	getchar();
	return 0;
}

RBTree.h(头文件)

#pragma once

enum Color
{
	RED,BLACK,};

template<class K,class V>
struct RBTreeNode
{
	K _key;
	V _value;

	RBTreeNode<K,V>* _left;
	RBTreeNode<K,V>* _right;
	RBTreeNode<K,V>* _parent;

	Color _color;

	RBTreeNode(const K& key,const V& value)
		:_key(key),_value(value),_left(NULL),_right(NULL),_parent(NULL),_color(RED)
	{}
};

template<class K,class V>
class RBTree
{
	typedef RBTreeNode<K,V> Node;
public:
	RBTree()
		:_root(NULL)
	{}
	bool Insert(const K& key,const V& value)
	{
		if(_root == NULL)
		{
			_root = new Node(key,value);
			_root->_color = BLACK;
			return true;
		}

		Node* cur = _root;
		Node* parent = NULL;
		while(cur)
		{
			if(cur->_key < key)
			{
				parent = cur;
				cur = cur->_right;
			}
			else if(cur->_key > key)
			{
				parent = cur;
				cur = cur->_left;
			}
			else
			{
				return false;
			}
		}
		cur = new Node(key,value);
		if(parent->_key > key)
		{
			parent->_left = cur;
			cur->_parent = parent;
		}
		else if(parent->_key < key)
		{
			parent->_right = cur;
			cur->_parent = parent;
		}

		// 调整颜色
		while(cur != _root && parent->_color == RED)
		{
			Node* grandfather = parent->_parent;
			if (parent == grandfather->_left)
			{
				Node* uncle = grandfather->_right;
				// 情况1:cur为红，p为红，g为黑，u存在且为红
				if (uncle && uncle->_color == RED)
				{
					parent->_color = BLACK;
					uncle->_color = BLACK;
					grandfather->_color = RED;

					cur = grandfather;
					parent = cur->_parent;
				}
				//叔叔不存在，存在且为黑色
				else if((uncle == NULL) || (uncle != NULL && uncle->_color == BLACK))
				{
					//情况3:cur为红，p为红，g为黑，u不存在/u为黑(cur为parent的右孩子)
					//情况2:cur为红，p为红，g为黑，u不存在/u为黑(cur为parent的左孩子)
					if (parent->_right == cur)
					{
						RotateL(parent);
						swap(cur,parent);
					}
					parent->_color = BLACK;
					grandfather->_color = RED;
					RotateR(grandfather);
					break;
				}
			}
			else // parent == grandfather->_right
			{
				Node* uncle = grandfather->_left;
				if (uncle && uncle->_color == RED)//叔叔为左，且为红色
				{
					parent->_color = uncle->_color = BLACK;
					grandfather->_color = RED;

					cur = grandfather;
					parent = cur->_parent;
				}
				//叔叔不存在，存在且为黑色
				else if((uncle == NULL) || (uncle != NULL && uncle->_color == BLACK))
				{
					if (cur == parent->_left)
					{
						RotateR(parent);
						swap(cur,parent);
					}
					parent->_color = BLACK;
					grandfather->_color = RED;
					RotateL(grandfather);
					break;
				}
			}
		}
		_root->_color = BLACK;//根节点始终是黑色
		return true;
	}

	void RotateL(Node* parent)
	{
		Node* subR = parent->_right;
		Node* subRL = subR->_left;
		parent->_right = subRL;
		if(subRL)
		{
			subRL->_parent = parent;
		}
		Node* ppNode = parent->_parent;
		subR->_left = parent;
		parent->_parent = subR;

		if(ppNode == NULL)
		{
			_root = subR;
			subR->_parent = NULL;
		}
		else
		{
			if(ppNode->_left == parent)
			{
				ppNode->_left = subR;
				subR->_parent = ppNode;
			}
			else
			{
				ppNode->_right = subR;
				subR->_parent = ppNode;
			}
		}
	}

	void RotateR(Node* parent)
	{
		Node* subL = parent->_left;
		Node* subLR = subL->_right;
		parent->_left = subLR;
		if(subLR)
		{
			subLR->_parent = parent;
		}

		Node* ppNode = parent->_parent;
		subL->_right = parent;
		parent->_parent = subL;

		if(ppNode ==  NULL)
		{
			_root = subL;
			subL->_parent = NULL;
		}
		else
		{
			if(ppNode->_left == parent)
			{
				ppNode->_left = subL;
			}
			else
			{
				ppNode->_right = subL;
			}
			subL->_parent = ppNode;
		}
	}

	void InOrder()
	{
		return _InOrder(_root);
		cout<<endl;
	}

	bool IsBlance()
	{
		//1.检查根节点是否为黑色节点
		//2.检查每条路径上黑色节点的个数是否相等
		//3.检查是否有连续的红色节点
		int BlackCount = 0;
		Node* cur = _root;
		while(cur)
		{
			if(cur->_color == BLACK)
			{
				BlackCount++;
			}
			cur = cur->_left;
		}
		int curBlackCount = 0;
		return _IsBlance(_root,BlackCount,curBlackCount);
	}

	Node* Find(const K& key)
	{
		return _Find(_root,key);
	}
	bool Remove(const K& key)
	{}

protected:
	Node* _Find(Node* root,const K& key)
	{
		if(_root == NULL)
		{
			return NULL;
		}
		while(root)
		{
			if(root->_key == key)
			{
				cout<<"Find"<<root->_key<<":"<<root->_value<<endl;
				return root;
			}
			else if(root->_key > key)
				root = root->_left;
			else
				root = root->_right;
		}
		
		cout<<"没有找到该节点"<<endl;
		return NULL;
		
	}

	bool _IsBlance(Node* root,int BlackCount,int curBlackCount)
	{
		if(root == NULL)
		{
			return true;
		}
		//1.检查根节点是否黑色节点
		if(_root->_color == RED)
		{
			return false;
		}
		//3.是否有连续的红色节点
		if(root->_color == BLACK)
		{
			curBlackCount++;
		}
		else
		{
			if(root->_parent && root->_parent->_color == RED)
			{
				cout<<"有连续的红色节点:"<<root->_key<<endl;
				return false;
			}
		}
		//2.检查每条路径上黑色节点的个数
		if(root->_left == NULL && root->_right == NULL)
		{
			if(BlackCount == curBlackCount)
			{
				return true;
			}
			else
			{
				cout<<"黑色节点数量不相等"<<root->_key<<endl;
				return false;
			}
		}
		return _IsBlance(root->_left,curBlackCount)
			&& _IsBlance(root->_right,curBlackCount);

	}

	void _InOrder(Node* root)
	{
		if(root == NULL)
		{
			return;
		}
		_InOrder(root->_left);
		cout<<root->_key<<" ";
		_InOrder(root->_right);
	}
protected:
	Node* _root;
};


void TestTree()
{
	RBTree<int,int> tree;
	int array[]={16,3,7,11,9,26,18,14,15};
	for(size_t i = 0; i < sizeof(array)/sizeof(array[0]); ++i)
	{
		tree.Insert(array[i],i);
	}
	tree.InOrder();
	cout<<endl;
	cout<<"IsBlance?"<<tree.IsBlance()<<endl;;
	tree.Find(18);
	tree.Find(13);
}

插入的5种情形：

情形1：该树为空树，直接插入根结点的位置，违反性质1，把节点颜色由红改为黑即可。

 
情形2：插入节点N的父节点P为黑色，不违反任何性质，无需做任何修改。
情形3：cur为红，p为红，g为黑，u存在且为红
操作：则将p,u改为黑，g改为红，然后把g当成cur，继续向上调整。

情形4：cur为红，p为红，g为黑，u不存在/u为黑
操作：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
情形5：cur为红，p为红，g为黑，u不存在/u为黑
操作：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4