二分搜索树

Posted lzb0803

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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 *root;
    int count;

public:
    BST(){
        root = NULL;
        count = 0;
    }
    ~BST(){
        // TODO: ~BST()
    }

    int size(){
        return count;
    }

    bool isEmpty(){
        return count == 0;
    }
};

 

插入新的节点:

publicvoid insert(Key key, Value value){
        root = insert(root, key, value);
    }

private:
    // 向以node为根的二叉搜索树中,插入节点(key, value)
    // 返回插入新节点后的二叉搜索树的根
    Node* insert(Node *node, Key key, Value value){

        if( node == NULL ){
            count ++;
            return new Node(key, value);
        }

        if( key == node->key )
            node->value = value;
        else if( key < node->key )
            node->left = insert( node->left , key, value);
        else    // key > node->key
            node->right = insert( node->right, key, value);

        return node;
    }
};

 

是否包含有键值为key的节点:

publicbool contain(Key key){
        return contain(root, key);
    }

private// 查看以node为根的二叉搜索树中是否包含键值为key的节点
    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 );
    }

 

查找:

public:
    Value* search(Key key){
        return search( root , key );
    }

private// 在以node为根的二叉搜索树中查找key所对应的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 );
    }

 

前序遍历:

public// 前序遍历
    void preOrder(){
        preOrder(root);
    }

private// 对以node为根的二叉搜索树进行前序遍历
    void preOrder(Node* node){

        if( node != NULL ){
            cout<<node->key<<endl;
            preOrder(node->left);
            preOrder(node->right);
        }
    }

中序遍历:

public// 中序遍历
    void inOrder(){
        inOrder(root);
    }


private// 对以node为根的二叉搜索树进行中序遍历
    void inOrder(Node* node){

        if( node != NULL ){
            inOrder(node->left);
            cout<<node->key<<endl;
            inOrder(node->right);
        }
    }

后序遍历:

public// 后序遍历
    void postOrder(){
        postOrder(root);
    }

private// 对以node为根的二叉搜索树进行后序遍历
    void postOrder(Node* node){

        if( node != NULL ){
            postOrder(node->left);
            postOrder(node->right);
            cout<<node->key<<endl;
        }
    }

 

析构函数:

public~BST(){
        destroy( root );
    }

privatevoid destroy(Node* node){

        if( node != NULL ){
            destroy( node->left );
            destroy( node->right );

            delete node;
            count --;
        }
    }

层序遍历:

public// 层序遍历
    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 );
        }
    }

 

最小键值:

public// 寻找最小的键值
    Key minimum(){
        assert( count != 0 );
        Node* minNode = minimum( root );
        return minNode->key;
    }

private// 在以node为根的二叉搜索树中,返回最小键值的节点
    Node* minimum(Node* node){
        if( node->left == NULL )
            return node;

        return minimum(node->left);
    }

最大键值:

public// 寻找最大的键值
    Key maximum(){
        assert( count != 0 );
        Node* maxNode = maximum(root);
        return maxNode->key;
    }

private// 在以node为根的二叉搜索树中,返回最大键值的节点
    Node* maximum(Node* node){
        if( node->right == NULL )
            return node;

        return maximum(node->right);
    }

删除最小节点:

public// 从二叉树中删除最小值所在节点
    void removeMin(){
        if( root )
            root = removeMin( root );
    }
private// 删除掉以node为根的二分搜索树中的最小节点
    // 返回删除节点后新的二分搜索树的根
    Node* removeMin(Node* node){

        if( node->left == NULL ){

            Node* rightNode = node->right;
            delete node;
            count --;
            return rightNode;
        }

        node->left = removeMin(node->left);
        return node;
    }

删除最大节点:

public// 从二叉树中删除最大值所在节点
    void removeMax(){
        if( root )
            root = removeMax( root );
    }

private// 删除掉以node为根的二分搜索树中的最大节点
    // 返回删除节点后新的二分搜索树的根
    Node* removeMax(Node* node){

        if( node->right == NULL ){

            Node* leftNode = node->left;
            delete node;
            count --;
            return leftNode;
        }

        node->right = removeMax(node->right);
        return node;
    }

删除任意节点:

public:
    Node(Node *node){
            this->key = node->key;
            this->value = node->value;
            this->left = node->left;
            this->right = node->right;
        }

 

public// 从二叉树中删除键值为key的节点
    void remove(Key key){
        root = remove(root, key);
    }

private// 删除掉以node为根的二分搜索树中键值为key的节点
    // 返回删除节点后新的二分搜索树的根
    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;
        }
    }

 

 

                      

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