错误C2679二进制＆＃39; =＆＃39;：未找到哪个运算符采用类型＆＃39; int＆＃39;的右侧操作数。 （或者没有可接受的转换）

Question

我编写了一个程序实现树并首先遍历这个树的广度。该程序包括3个文件：queue.h，tree.h和main.cpp ......这是queue.h

#pragma once
#include <iostream>

template <class T>
struct L1node {
    T data;
    L1node* next = NULL;
};

template <class T>
class L1queue
{
    L1node<T>* pHead, pTail;
public:
    L1queue() :pHead(NULL), pTail(NULL) {}
    ~L1queue() {
        if (pHead) {
            while (pHead->next) {
                L1node<T>* p = pHead;
                pHead = pHead->next;
                delete p;
            }
            delete pHead;
        }
        return 0;
    }
    bool isEmpty() {
        return pHead == NULL:
    }
    T& head() {
        if (isEmpty()) throw - 1;
        return pHead->data;
    }
    void enqueue(T& a) {
        L1node<T> p = new L1node<T>(a);
        if (isEmpty()) {
            pHead = pTail = p;
        }
        else {
            pTail->next = p;
            pTail = p;
        }
    }
    void dequeue() {
        if (isEmpty()) return;
        L1node<T>* p = pHead;
        pHead = pHead->next;
        delete p;
        if (pHead == NULL) pTail = NULL;   //Error here
    }
};

这是tree.h

#include <iostream>
#include "queue.h"
using namespace std;

class treeNode {
public:
    int data;
    treeNode* left = NULL;
    treeNode* right = NULL;
};
void printLeavesBFT(treeNode* root) {
    L1queue<treeNode*> q;
    while (!q.isEmpty()) {
        treeNode* p = q.head(); cout << p->data << " ";
        q.dequeue();
        if(p->left) q.enqueue(root->left);
        if(p->right) q.enqueue(root->right);
    }
}

我遇到了出队功能错误。我不知道原因。你能帮帮我吗？

Answer 1

pTail被声明为L1node<T>，而您打算将其声明为L1node<T>*。然后，pTail = NULL毫无意义。

当您编写L1node<T>* pHead, pTail;时，只有pHead是指针，pTail是对象。

替换为

L1node<T> *pHead, *pTail;

或者：

L1node<T>* pHead;
L1node<T>* pTail;