【C++】TreeNode 类的析构函数有问题，tn1的析构报错，不知怎么修改，求告知[大哭]

Question

#include <vector>
#include <iostream>
#include <fstream>
#include <algorithm>
#include <string>
#include <memory>

using std::cin;
using std::cout;
using std::endl;
using std::string;
using std::vector;
using std::shared_ptr;

class TreeNode {

public:

    TreeNode() :count(0), left(nullptr), right(nullptr), num(new std::size_t(1)) {}
    TreeNode(const string& str, int c) :value(str),count(c), left(nullptr), right(nullptr), num(new std::size_t(1)) {}
    TreeNode(const string& str, int c, TreeNode& lnode, TreeNode& rnode) :value(str), count(c), left(&lnode), right(&rnode), num(new std::size_t(1)) {}
    TreeNode(const TreeNode& tn) :value(tn.value), count(tn.count), left(tn.left), right(tn.right) { ++* num; }

    ~TreeNode() {
        if (-- * num == 0) {
            delete left;
            delete right;
            delete num;
        }
    }

    TreeNode& operator=(const TreeNode& tn) {

        ++* tn.num;
        if (-- * num == 0) {
            delete left;
            delete right;
            delete num;
        }
        value = tn.value;
        count = tn.count;
        left = tn.left;
        right = tn.right;

        return *this;
    }

private:
    string value;
    int count;
    TreeNode* left;
    TreeNode* right;
        //模拟引用计数，*num为0则销毁对象
    std::size_t* num;
};

void testTreeNode() {
    TreeNode tn2("World", 1025), tn3("!",1026);
    TreeNode tn1("Hello", 1024, tn2, tn3);

        //tn1 = tn3;

}

int main(){
    testTreeNode();
    cout << "end...\n";
    return 0;
}

Answer 1

晕，刚才回答完，发现你删除了
tn2,tn3先析构掉了，相当于tn1的left和right被析构掉了，然后tn1的析构的时候你再析构left和right就会报错啊
原因你只是进行指针的赋值，没有进行深拷贝
在构造函数中，left和right 创建新对象，然后将tn2和tn3的值赋值给left和right新建对象的值，这样就互不相关了