#include <iostream>
#include <queue>
#include <cassert>
using namespace std;
template <typename Key, typename Value>
class BST{
private:
struct Node{
Key key;
Value value;
Node *left;
Node *right;
Node(Key key, Value value){
this->key = key;
this->value = value;
this->left = this->right = NULL;
}
Node(Node *node){
this->key = node->key;
this->value = node->value;
this->left = node->left;
this->right = node->right;
}
};
Node *root;
int count;
public:
BST(){
root = NULL;
count = 0;
}
~BST(){
destroy( root );
}
int size(){
return count;
}
bool isEmpty(){
return count == 0;
}
int main() {
srand(time(NULL));
BST<int,int> bst = BST<int,int>();
int n = 10000;
for( int i = 0 ; i < n ; i ++ ){
int key = rand()%n;
// 為了后續(xù)測(cè)試方便,這里value值取和key值一樣
int value = key;
//cout<<key<<" ";
bst.insert(key,value);
}
// test remove
// remove elements in random order
int order[n];
for( int i = 0 ; i < n ; i ++ )
order[i] = i;
shuffle( order , n );
for( int i = 0 ; i < n ; i ++ )
if( bst.contain( order[i] )){
bst.remove( order[i] );
cout<<"After remove "<<order[i]<<" size = "<<bst.size()<<endl;
}
return 0;
}
1. 插入insert
public:
void insert(Key key, Value value){
root = insert(root, key, value);
}
private:
// 向以node為根的二叉搜索樹中,插入節(jié)點(diǎn)(key, value)
// 返回插入新節(jié)點(diǎn)后的二叉搜索樹的根
Node* insert(Node *node, Key key, Value value){
if( node == NULL ){ // 如果節(jié)點(diǎn)為空,就在此節(jié)點(diǎn)處加入x信息
count ++;
return new Node(key, value);
}
if( key == node->key )
node->value = value; //如果相等,就把頻率加1
else if( key < node->key )
node->left = insert( node->left , key, value); // 如果x小于節(jié)點(diǎn)的值,就繼續(xù)在節(jié)點(diǎn)的左子樹中插入x
else // key > node->key
node->right = insert( node->right, key, value); // 如果x大于節(jié)點(diǎn)的值,就繼續(xù)在節(jié)點(diǎn)的右子樹中插入x
return node;
}
2. 包含contain
bool contain(Key key){
return contain(root, key);
}
// 查看以node為根的二叉搜索樹中是否包含鍵值為key的節(jié)點(diǎn)
bool contain(Node* node, Key key){
if( node == NULL )
return false;
if( key == node->key )
return true;
else if( key < node->key )
return contain( node->left , key );
else // key > node->key
return contain( node->right , key );
}
3. 搜索search
Value* search(Key key){
return search( root , key );
}
// 在以node為根的二叉搜索樹中查找key所對(duì)應(yīng)的value
Value* search(Node* node, Key key){
if( node == NULL )
return NULL;
if( key == node->key )
return &(node->value);
else if( key < node->key )
return search( node->left , key );
else // key > node->key
return search( node->right, key );
}
4. 遍歷
void preOrder(){
preOrder(root);
}
// 對(duì)以node為根的二叉搜索樹進(jìn)行前序遍歷
void preOrder(Node* node){
if( node != NULL ){
cout<<node->key<<endl;
preOrder(node->left);
preOrder(node->right);
}
}
void inOrder(){
inOrder(root);
}
// 對(duì)以node為根的二叉搜索樹進(jìn)行中序遍歷
void inOrder(Node* node){
if( node != NULL ){
inOrder(node->left);
cout<<node->key<<endl;
inOrder(node->right);
}
}
void postOrder(){
postOrder(root);
}
// 對(duì)以node為根的二叉搜索樹進(jìn)行后序遍歷
void postOrder(Node* node){
if( node != NULL ){
postOrder(node->left);
postOrder(node->right);
cout<<node->key<<endl;
}
}
void levelOrder(){
queue<Node*> q;
q.push(root);
while( !q.empty() ){
Node *node = q.front();
q.pop();
cout<<node->key<<endl;
if( node->left )
q.push( node->left );
if( node->right )
q.push( node->right );
}
}
5. 尋找最小/大鍵值
Key minimum(){
assert( count != 0 );
Node* minNode = minimum( root );
return minNode->key;
}
// 在以node為根的二叉搜索樹中,返回最小鍵值的節(jié)點(diǎn)
Node* minimum(Node* node){
if( node->left == NULL )
return node;
return minimum(node->left);
}
Key maximum(){
assert( count != 0 );
Node* maxNode = maximum(root);
return maxNode->key;
}
// 在以node為根的二叉搜索樹中,返回最大鍵值的節(jié)點(diǎn)
Node* maximum(Node* node){
if( node->right == NULL )
return node;
return maximum(node->right);
}
6. 從二叉樹中刪除最小/大值所在節(jié)點(diǎn)
void removeMin(){
if( root )
root = removeMin( root );
}
// 刪除掉以node為根的二分搜索樹中的最小節(jié)點(diǎn)
// 返回刪除節(jié)點(diǎn)后新的二分搜索樹的根
Node* removeMin(Node* node){
if( node->left == NULL ){
Node* rightNode = node->right;
delete node;
count --;
return rightNode;
}
node->left = removeMin(node->left);
return node;
}
void removeMax(){
if( root )
root = removeMax( root );
}
// 刪除掉以node為根的二分搜索樹中的最大節(jié)點(diǎn)
// 返回刪除節(jié)點(diǎn)后新的二分搜索樹的根
Node* removeMax(Node* node){
if( node->right == NULL ){
Node* leftNode = node->left;
delete node;
count --;
return leftNode;
}
node->right = removeMax(node->right);
return node;
}
7. 刪除remove
void remove(Key key){
root = remove(root, key);
}
// 刪除掉以node為根的二分搜索樹中鍵值為key的節(jié)點(diǎn)
// 返回刪除節(jié)點(diǎn)后新的二分搜索樹的根
Node* remove(Node* node, Key key){
if( node == NULL )
return NULL;
if( key < node->key ){
node->left = remove( node->left , key );
return node;
}
else if( key > node->key ){
node->right = remove( node->right, key );
return node;
}
else{ // key == node->key
if( node->left == NULL ){
Node *rightNode = node->right;
delete node;
count --;
return rightNode;
}
if( node->right == NULL ){
Node *leftNode = node->left;
delete node;
count--;
return leftNode;
}
// node->left != NULL && node->right != NULL
Node *successor = new Node(minimum(node->right));
count ++;
successor->right = removeMin(node->right);
successor->left = node->left;
delete node;
count --;
return successor;
}
}
