上篇部落格介紹了怎樣刪除二分搜尋樹的最大值和最小值節點，下面介紹怎樣刪除樹的任意一個節點
上篇刪除最大值節點的操作，其實刪除的節點要麼沒有左右子節點，要麼只可能有左節點，
同樣，刪除最小值節點的操作，其實刪除的節點要麼沒有左右子節點，要麼只可能有右節點

（1）刪除只有左節點的節點，如刪除節點58，只需要將58刪除掉，然後將其左節點50替換其位置即可

（2）刪除只有右節點的節點，如刪除節點58，只需要將58刪除掉，然後將其右節點60替換其位置即可

（3）刪除既有左節點又有右節點的節點是最複雜的操作，在1962年，Hibbard提出了一種刪除方法，稱為 Habbard Deletion，如下圖所示，如果需要刪除既有左節點又有右節點的節點58（稱為d），那麼需要找出58的右節點中最小的那個節點s，那麼s就是節點59，將節點59放到58的位置即可

#include <iostream>
#include <vector>
#include <string>
#include <ctime>  //time()函式
#include <queue>
#include <cassert>
#include "FileOps.h"
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(){
        // TODO: ~BST()
        destroy(root);
    }

    int size(){
        return count;
    }

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

    void insert(Key key, Value value){

        root = insert(root, key, value);
    }

    bool contain(Key key){

        return contain(root, key);
    }

    Value *search(Key key){

        return search(root, key);
    }

    //前序遍歷
    void preOrder(){
        preOrder(root);
    }

    //中序遍歷
    void inOrder(){
        inOrder(root);
    }

    //後序遍歷
    void postOrder(){
        postOrder(root);
    }

    //層序遍歷
    void levelOrder(){

        quene<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);
        }
    }

    //尋找最小的鍵值
    Key minimum(){
        assert(count != 0);
        Node* minNode = minimum(root);
        return minNode->key;
    }

    //尋找最大的鍵值
    Key maximum(){
        assert(count != 0);
        Node* maxNode = maximum(root);
        return maxNode->key;
    }

    //從二叉樹中刪除最小值所在節點
    void removeMin(){
        if (root)
            root = removeMin(root);
    }

    //從二叉樹中刪除最大值所在節點
    void removeMax(){
        if (root)
            root = removeMax(root);
    }

    //從二叉樹中刪除鍵值為key的節點
    void remove(Key key){
        root = remove(root, key);
    }



private:
    //向以node為根的二叉搜尋樹中，插入節點(key, value)
    //返回插入新節點後的二叉搜尋樹的根
    //遞迴插入
    Node* insert(Node *node, Key key, Value value){

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

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

        return node;
    }

    //查詢以node為根的二叉搜尋樹中是否包含鍵值為key的節點
    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
            return contain(node->right, key);
    }

    //在以node為根的二叉搜尋樹中查詢key對應的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
            return search(node->right, key);
    }

    //對以node為根的二叉搜尋樹進行前序遍歷
    void preOrder(Node* node){
        if (node != NULL){
            cout<<node->key<<endl;
            preOrder(node->left);
            preOrder(node->right);
        }
    }

    //對以node為根的二叉搜尋樹進行中序遍歷
    void inOrder(Node *node){
        if (node != NULL){
            inOrder(node->left);
            cout<<node->key<<endl;
            inOrder(node->right);

        }
    }

    //對以node為根的二叉搜尋樹進行後序遍歷
    void postOrder(Node *node){
        if(node != NULL){
            postOrder(node->left);
            postOrder(node->right);
            cout<<node->key<<endl;
        }
    }

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

            delete node;
            count--;
        }
    }

    //在以node為根的二叉搜尋樹中，返回最小鍵值的節點
    Node* minimum(Node *node){
        if (node->left == NULL)
            return node;

        return minimum(node->left);
    }

    //在以node為根的二叉搜尋樹中，返回最大鍵值的節點
    Node* maximum(Node *node){
        if (node->right == NULL)
            return node;

        return maximum(node->right);
    }

    //刪除掉以node為根的二分搜尋樹中的最小節點
    //返回刪除節點後新的二分搜尋樹的根
    Node* removeMin(Node* node){

        if (node->left == NULL){

            Node *rightNode = node->right;
            delete node;
            count--;
            return rightNode;
        }
        node->left = removeMin(node->left);
        return node;
    }

    //刪除掉以node為根的二分搜尋樹中的最大節點
    //返回刪除節點後新的二分搜尋樹的根
    Node* removeMax(Node* node){

        if (node->right == NULL){

            Node *leftNode = node->left;
            delete node;
            count--;
            return leftNode;
        }
        node->right = removeMin(node->right);
        return node;
    }

    //刪除掉以node為根的二分搜尋樹中鍵值為key的節點
    //返回刪除節點後新的二分搜尋樹的根
    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 *delNode = node;
            Node *successor = new Node(minimum(node->right));
            count++;

            successor->right = removeMin(node->right);
            successor->left = node->left;

            //delete delNode;
            delete node;
            count--;

            return successor;
        }
    }

};

上面刪除既有左節點又有右節點的節點時，使用的是該節點後驅59替代該節點，其實也可以用該節點的前驅53替代該節點。如下圖所示