C++ 将第二个元素添加到 BST - 段错误

Question

所以我根据类里面的例子构建了一个 BST。我的同行的树都工作正常，但是，当我添加第二个元素时，我遇到了段错误。我已经隔离了来自这一行的错误 - 'if(current->left == NULL){' 我不确定它为什么会导致段错误以及如何避免它。我将留下相关的 insertNode 代码以及 Node 类，因为我觉得那里可能有问题。

template <class T>
void BST<T>::insertNode(T value){
  TreeNode<T> *node = new TreeNode<T>(value);


  if(isEmpty()){
    cout << "Is root" << endl;
    root = node;
  }else{
    cout << "Is not root" << endl;
    TreeNode<T> *parent = NULL;
    TreeNode<T> *current = root;
    cout << "Starting Loop" << endl;
    while(true){
      parent = current;

      if(value < current->key){
        cout << "less tahn" << endl;
        //Left
        current = current->left;
        cout <<"left" << endl;
        if(current == NULL){
          //we found our location/insertion point
          cout << "Found Spot" << endl;
          parent->left = node;
          break;
        }
        cout << "Not NULL" << endl;
      }
      else {
        //Right
        cout << "Right" << endl;
        current = current->right;
        if(current == NULL){
          //we found our location/insertion point
          cout << "Found Spot" << endl;
          parent->right = node;
          break;
        }
      }
    }
  }
}

还有 treeNode 类 -

template <class T>
class TreeNode{
  public:
    TreeNode();
    TreeNode(T k);
    ~TreeNode();

    T key;
    TreeNode *left;
    TreeNode *right;
};
template <class T>
TreeNode<T>::TreeNode(){
  left = NULL;
  right = NULL;
}

template <class T>
TreeNode<T>::TreeNode(T k){
  left = NULL;
  right = NULL;
  key = k;
}

Answer 1

这是建立在 SHR 所说的基础上的。当您执行 current = current->left 行时，有可能current->left上一个电流是 null .
这意味着当您执行检查 if (current->left == NULL) ，会导致段错误，因为没有left .
您可以将条件重写为:

if (current && !current->left){
...
}

通过这样做，您首先检查是否 current不为空，如果为空，那么它才会查看节点的左指针。或者您也可以事先查看 if (current) .
检查 NULL 也完全足够并鼓励通过执行 if (node)而不是做 if (node == NULL) .这解释了原因: Checking for NULL pointer in C/C++