在C++中實現鏈表節點的反轉可以通過迭代或遞歸兩種方法來實現。以下是其中一種方法的示例代碼：

#include <iostream>

struct ListNode {
    int val;
    ListNode* next;
    ListNode(int x) : val(x), next(NULL) {}
};

ListNode* reverseList(ListNode* head) {
    ListNode* prev = NULL;
    ListNode* curr = head;
    while (curr != NULL) {
        ListNode* next = curr->next;
        curr->next = prev;
        prev = curr;
        curr = next;
    }
    return prev;
}

void printList(ListNode* head) {
    while (head != NULL) {
        std::cout << head->val << " ";
        head = head->next;
    }
    std::cout << std::endl;
}

int main() {
    ListNode* head = new ListNode(1);
    head->next = new ListNode(2);
    head->next->next = new ListNode(3);
    head->next->next->next = new ListNode(4);

    std::cout << "Original list: ";
    printList(head);

    head = reverseList(head);

    std::cout << "Reversed list: ";
    printList(head);

    return 0;
}

該示例代碼中定義了一個節點結構體ListNode，并實現了reverseList函數來反轉鏈表。通過調用reverseList函數并打印結果，可以看到鏈表節點的順序被成功反轉。