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  • 二叉查找树

      1 #include <iostream>
      2 
      3 using namespace std;
      4 
      5 template <typename K, typename V>
      6 class BST
      7 {
      8     private:
      9         struct node    // 类内定义类、结构体
     10         {
     11             K key;
     12             V value;
     13             node *left = nullptr;
     14             node *right = nullptr; 
     15         };
     16         node * root = nullptr;
     17         node * putB(node *rootp, K key, V value);   // 注意不能把这个函数放在公有区,因为node是一个私有变量,如果外部调用putB会出问题,且只有成员函数会调用这个函数,没必要声明为公有的就不要声明为公有
     18     public:
     19         BST(/* args */);
     20         void putA(K key, V value);
     21         void putB(K key, V value);
     22         /* void putB(node *&rootp, K key, V value); */  //类内定义结构体不用加作用域限定:  BST<K, V>::node
     23         
     24         V get(K key);
     25         V getR(node *rootp, K key);     //对于结构相似且头尾相接的数据结构(如这里的二叉树:每个节点都可以是根节点,每个根节点下都有左右节点),要善于利用递归
     26         bool max(K &key, V &value);
     27         bool min(K &key, V &value);
     28         int rank(K key);
     29         bool remove(K key);
     30         int howMany();
     31         void show();
     32         void showR(node *p);
     33         void free(node *p);
     34         ~BST(); 
     35 };
     36 
     37 template <typename K, typename V>
     38 BST<K, V>::BST(/* args */)
     39 {
     40     //did nothing
     41 }
     42 
     43 template <typename K, typename V>
     44 V BST<K, V>::get(K key)
     45 {
     46     V val = NULL;
     47     val = getR(root, key);
     48     if(val != NULL)    //warning:val == 0  is equal to val == NULL
     49     {
     50         cout << "key: " << key << "	 val: " << val << endl;
     51         return val;
     52     }
     53 }
     54 
     55 template <typename K, typename V>
     56 V BST<K, V>::getR(node *rootp, K key)
     57 {
     58     if(rootp == nullptr)
     59     {
     60         cout << "can't find key[" << key << "]." << endl;
     61         return NULL;
     62     }
     63         
     64     if(rootp->key == key)
     65         return rootp->value;
     66     else if(rootp->key < key)
     67     {
     68         return getR(rootp->right, key);
     69     }
     70     else 
     71         return getR(rootp->left, key);
     72 }
     73 
     74 /* 循环查找 */
     75 template <typename K, typename V>
     76 void BST<K, V>::putA(K key, V value)
     77 {
     78     node **temp = &root;    //使用指针的指针,因为下面要给指针分配内存
     79     while((*temp) != nullptr)
     80     {
     81         if((*temp)->key == key)
     82         {
     83             (*temp)->key = key;
     84             (*temp)->value = value;
     85             return;
     86         }
     87         if((*temp)->key > key)
     88             temp = &((*temp)->left);
     89         else 
     90             temp = &((*temp)->right);
     91     }
     92     *temp = (node *)new node;
     93     (*temp)->key = key;
     94     (*temp)->value = value;
     95 }
     96 
     97 template <typename K, typename V>
     98 void BST<K, V>::putB(K key, V value)
     99 {
    100     /* putB(root, key, value); */
    101     root = putB(root, key, value);
    102 }
    103 
    104 /* 递归查找 */
    105 /* 
    106 template <typename K, typename V>
    107 void BST<K, V>::putB(node *&rootp, K key, V value)   //指针的引用
    108 {
    109     if(rootp == nullptr)
    110     {
    111         rootp = new node;
    112         rootp->key = key;
    113         rootp->value = value;
    114         return;
    115     }
    116     if(rootp->key == key)
    117     {
    118         rootp->value = value;
    119         return;
    120     }
    121     else if(rootp->key > key)
    122         return putB(rootp->left, key, value);
    123     else
    124         return putB(rootp->right, key, value);
    125     
    126 }
    127  */
    128 
    129 /* 不加typename的话会提示错误: need 'typename' before 'BST<K, V>::node' because 'BST<K, V>' is a dependent scope,原因:c++primer P593 */
    130 template <typename K, typename V>
    131 typename BST<K, V>::node * BST<K, V>::putB(node *rootp, K key, V value)   
    132 {
    133     if(rootp == nullptr)
    134     {
    135         rootp = new node;
    136         rootp->key = key;
    137         rootp->value = value;
    138         return rootp;
    139     }
    140     if(rootp->key == key)
    141     {
    142         rootp->value = value;
    143     }
    144     else if(rootp->key > key)
    145         rootp->left =  putB(rootp->left, key, value);
    146     else
    147         rootp->right = putB(rootp->right, key, value);
    148 
    149     return rootp;
    150 }
    151 
    152 template <typename K, typename V>
    153 void BST<K, V>::show()
    154 {
    155     showR(root);
    156 }
    157 
    158 template <typename K, typename V>
    159 void BST<K, V>::showR(node *p)
    160 {
    161     if(p->left != nullptr)
    162         showR(p->left);
    163     cout << "key: " << p->key << "	value: " << p->value << endl;
    164     if(p->right != nullptr)
    165         showR(p->right);
    166 }
    167 
    168 template <typename K, typename V>
    169 void BST<K, V>::free(node *p)
    170 {
    171     if(p == nullptr)
    172         return;
    173     free(p->left);
    174     free(p->right);
    175     delete p;
    176 }
    177 
    178 template <typename K, typename V>
    179 BST<K, V>::~BST()
    180 {
    181     free(root);
    182 }
    183 
    184 int main()
    185 {
    186     BST<char, int> container;
    187     cout << "before: " << endl;
    188     for(int i=0; i<22; ++i)
    189     {
    190         char k = 65 + rand() % 20;
    191         int v = rand() % 100;
    192         cout << "k: " << k << "    v: " << v << endl;
    193         container.putB(k, v);
    194         //or container.putA(k, v);
    195     }
    196     
    197     container.show();
    198 
    199     char token;
    200     cout << "enter key you want to find(0 to exit):"; 
    201     cin >> token;
    202     while(token != '0')
    203     {
    204         container.get(token);
    205         cout << "enter key you want to find(0 to exit):";
    206         cin >> token;
    207     }
    208     
    209     return 0;
    210 }
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  • 原文地址:https://www.cnblogs.com/tan-wm/p/14399091.html
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