0
点赞
收藏
分享

微信扫一扫

GPT-4o mini 来袭:开发者如何驾驭新一代AI模型?

 🌈个人主页:秦jh_-CSDN博客
🔥 系列专栏:https://blog.csdn.net/qinjh_/category_12575764.html?spm=1001.2014.3001.5482

 9efbcbc3d25747719da38c01b3fa9b4f.gif​ 

目录

前言

红黑树的概念

红黑树的性质

节点的定义

红黑树的插入操作 

 检测操作:

情况一: cur为红,p为红,g为黑,u存在且为红

情况二: cur为红,p为红,g为黑,u不存在/u存在且为黑 

Insert 代码

 红黑树的验证

红黑树与AVL树的比较

 完整代码


前言

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。(不存在连续的红节点)
  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(每条路径都存在相同数量的黑色节点)  
  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)

节点的定义

 

红黑树的插入操作 

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
  2. 检测新节点插入后,红黑树的性质是否造到破坏 

 检测操作:

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

 

 

情况二: cur为红,p为红,g为黑,u不存在/u存在且为黑 

情况二的双旋情况:

Insert 代码

bool Insert(const pair<K, V>& kv)
{
if (_root == nullptr)
{
_root = new Node(kv);
_root->_col = BLACK; //根节点默认黑色
return true;
}

Node* cur = _root;
Node* parent = nullptr;
while (cur)
{
if (cur->_kv.first < kv.first)
{
parent = cur;
cur = cur->_right;
}
else if (cur->_kv.first > kv.first)
{
parent = cur;
cur = cur->_left;
}
else
{
return false;
}
}

cur = new Node(kv);
cur->_col = RED; //新增节点给红色
if (parent->_kv.first > kv.first)
{
parent->_left = cur;
}
else
{
parent->_right = cur;
}
cur->_parent = parent;

// 检测新节点插入后,红黑树的性质是否造到破坏
//父亲的颜色是黑色也结束
while (parent && parent->_col == RED)
{
//关键看叔叔
Node* grandfather = parent->_parent;
if (parent == grandfather->_left)
{
Node* uncle = grandfather->_right;
//如果叔叔存在也为红->变色即可
if (uncle && uncle->_col == RED)
{
parent->_col = uncle->_col = BLACK;
grandfather->_col = RED;

//继续往上处理
cur = grandfather;
parent = cur->_parent;
}
else //叔叔不存在,或者存在且为黑
{
if (cur == parent->_left)
{
// g
// p u
// c
//单旋
RotateR(grandfather);
parent->_col = BLACK;
grandfather->_col = RED;
}
else
{
// g
// p u
// c
//双旋
RotateL(parent);
RotateR(grandfather);
cur->_col = BLACK;
grandfather->_col = RED;
}
break;
}
}
else
{
Node* uncle = grandfather->_left;
//如果叔叔存在也为红->变色即可
if (uncle && uncle->_col == RED)
{
parent->_col = uncle->_col = BLACK;
grandfather->_col = RED;

//继续往上处理
cur = grandfather;
parent = cur->_parent;
}
else //叔叔不存在,或者存在且为黑
{
// g
// u p
// c
if (cur == parent->_right)
{
RotateL(grandfather);
parent->_col = BLACK;
grandfather->_col = RED;
}
else
{
// g
// u p
// c
RotateR(parent);
RotateL(grandfather);
cur->_col = BLACK;
grandfather->_col = RED;
}
break;
}
}
}
//始终保持根为黑
_root->_col = BLACK;

return true;
}

 红黑树的验证

 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

	bool IsBalance()
{
if (_root->_col == RED)
{
return false;
}

int refNum = 0; //取其中一条路径作为参考值
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)
{
//cout << blackNum << endl;
if (refNum != blackNum)
{
cout << "存在黑色节点数量不相等的路径" << endl;
return false;
}
return true;
}

if (root->_col == RED && root->_parent->_col == RED)
{
cout << root->_kv.first << "存在连续的红色节点" << endl;
return false;
}

if (root->_col == BLACK)
{
blackNum++;
}

return Check(root->_left,blackNum,refNum)
&& Check(root->_right,blackNum, refNum);
}

红黑树与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(logN),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。 

 完整代码

#pragma once 

enum Colour
{
RED,
BLACK
};

template<class K,class V>
struct RBTreeNode
{
RBTreeNode<K, V>* _left;
RBTreeNode<K, V>* _right;
RBTreeNode<K, V>* _parent;

pair<K, V> _kv;
Colour _col;

RBTreeNode(const pair<K, V>& kv)
:_left(nullptr)
, _right(nullptr)
, _parent(nullptr)
, _kv(kv)
, _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; //根节点默认黑色
return true;
}

Node* cur = _root;
Node* parent = nullptr;
while (cur)
{
if (cur->_kv.first < kv.first)
{
parent = cur;
cur = cur->_right;
}
else if (cur->_kv.first > kv.first)
{
parent = cur;
cur = cur->_left;
}
else
{
return false;
}
}

cur = new Node(kv);
cur->_col = RED; //新增节点给红色
if (parent->_kv.first > kv.first)
{
parent->_left = cur;
}
else
{
parent->_right = cur;
}
cur->_parent = parent;

// 检测新节点插入后,红黑树的性质是否造到破坏
//父亲的颜色是黑色也结束
while (parent && parent->_col == RED)
{
//关键看叔叔
Node* grandfather = parent->_parent;
if (parent == grandfather->_left)
{
Node* uncle = grandfather->_right;
//如果叔叔存在也为红->变色即可
if (uncle && uncle->_col == RED)
{
parent->_col = uncle->_col = BLACK;
grandfather->_col = RED;

//继续往上处理
cur = grandfather;
parent = cur->_parent;
}
else //叔叔不存在,或者存在且为黑
{
if (cur == parent->_left)
{
// g
// p u
// c
//单旋
RotateR(grandfather);
parent->_col = BLACK;
grandfather->_col = RED;
}
else
{
// g
// p u
// c
//双旋
RotateL(parent);
RotateR(grandfather);
cur->_col = BLACK;
grandfather->_col = RED;
}
break;
}
}
else
{
Node* uncle = grandfather->_left;
//如果叔叔存在也为红->变色即可
if (uncle && uncle->_col == RED)
{
parent->_col = uncle->_col = BLACK;
grandfather->_col = RED;

//继续往上处理
cur = grandfather;
parent = cur->_parent;
}
else //叔叔不存在,或者存在且为黑
{
// g
// u p
// c
if (cur == parent->_right)
{
RotateL(grandfather);
parent->_col = BLACK;
grandfather->_col = RED;
}
else
{
// g
// u p
// c
RotateR(parent);
RotateL(grandfather);
cur->_col = BLACK;
grandfather->_col = RED;
}
break;
}
}
}
//始终保持根为黑
_root->_col = BLACK;

return true;
}

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

parent->_left = subLR;
if (subLR) //节点可能为空
subLR->_parent = parent;

subL->_right = parent; //旧父节点变成subL的右节点

Node* ppNode = parent->_parent; //该不平衡节点可能不是根节点,所以要找到它的父节点
parent->_parent = subL;

if (parent == _root) //如果该节点是根节点
{
_root = subL;
_root->_parent = nullptr;
}
else //不平衡节点只是一棵子树
{
if (ppNode->_left == parent) //如果旧父节点等于爷爷节点的左节点,新父节点为爷爷节点的左节点
{
ppNode->_left = subL;
}
else
{
ppNode->_right = subL;
}
subL->_parent = ppNode; //新父节点指向爷爷节点。
}
}

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

parent->_right = subRL;
if (subRL)
subRL->_parent = parent;

subR->_left = parent;
Node* ppNode = parent->_parent;

parent->_parent = subR;

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

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

bool IsBalance()
{
if (_root->_col == RED)
{
return false;
}

int refNum = 0; //取其中一条路径作为参考值
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)
{
//cout << blackNum << endl;
if (refNum != blackNum)
{
cout << "存在黑色节点数量不相等的路径" << endl;
return false;
}
return true;
}

if (root->_col == RED && root->_parent->_col == RED)
{
cout << root->_kv.first << "存在连续的红色节点" << endl;
return false;
}

if (root->_col == BLACK)
{
blackNum++;
}

return Check(root->_left,blackNum,refNum)
&& Check(root->_right,blackNum, refNum);
}


void _InOrder(Node* root)
{
if (root == nullptr)
return;

_InOrder(root->_left);
cout << root->_kv.first << ":" << root->_kv.second << endl;
_InOrder(root->_right);
}

private:
Node* _root = nullptr;
size_t _size = 0;
};


void RBTreeTest1()
{
//int a[] = { 8, 3, 1, 10, 6, 4, 7, 14, 13 };
int a[] = { 4, 2, 6, 1, 3, 5, 15, 7, 16, 14,8, 3, 1, 10, 6, 4, 7, 14, 13 };
RBTree<int, int> t1;
for (auto e : a)
{
t1.Insert({ e,e });

//cout << "Insert:" << e << "->" << t1.IsBalance() << endl;
}

t1.InOrder();

cout << t1.IsBalance() << endl;
}
举报

相关推荐

0 条评论