剑指offer13 在O(1)时间删除链表的结点

把下一个节点的值直接赋值给要删除的节点，然后删除下一个节点。当这样做会有两个bad case：被删除的链表结点的下一个结点为空指针，如果链表只有一个结点。其实链表只有一个结点应该属于下一个结点为空指针的大类，但处理方式不同。整个方案就是下一个结点是否是空指针

 

void deleteNode(TreeNode* pListHead,TreeNode* pToBeDeleted){
    if(pListHead == NULL || pToBeDeleted == NULL)
        return;
    if(pToBeDeleted->next != NULL){
        TreeNode* pnext == pToBeDeleted->next;
        pToBeDeleted->val = pnext->val;
        pToBeDeleted->next = pnext->next;
        delete pnext;
        pnext = NULL;
    }
    else if(pListHead == pToBeDeleted){
        pListHead == NULL;
        delete pToBeDeleted;
        pToBeDeleted == NULL;
    }
    else{
        TreeNode* pNode = pListHead;
        while(pNode->next != pToBeDeleted){
            pNode = pNode->next;
        }
        pNode->next == NULL;
        delete pToBeDeleted;
        pToBeDeleted == NULL;
    }
    return;
}