C++ delete问题

1. delete删除的该指针指向的内存，指针变成一个野指针

2. delete指针后，指针不一定为NULL:

   //BanaryNode是一个自定义的类   
    BanaryNode* tree = new BanaryNode();
    delete tree;
    if(tree != NULL)
      std::cout<<"NOT NULL"<<std::endl;
    //最后输出为NOT NULL

3. 为了保证delete后指针安全，所以需要将指针赋为NULL

    BanaryNode* tree = new BanaryNode();
    delete tree;
    tree = NULL;

4. 在函数传递指针过程中，会出现这样的问题:

 void BanaryRemove(BanaryNode* tree)
 {
   if(tree == NULL)
     return;
   delete tree;
   tree = NULL;
 }

   BanaryNode* tree = new BanaryNode();
   BanaryRemove(tree);
   if(tree != NULL)
        std::cout<<"NOT NULL"<<std::endl;
   //最后输出为NOT NULL

因为 BanaryRemove(BanaryNode* tree)中传递的值tree的值，在函数中将 tree = NULL，只是将tree的副本赋值为NULL。

可通过引用传递解决问题，通过直接对tree的操作而不是tree的副本

  void BanaryRemove(BanaryNode* &tree)
  {
    if(tree == NULL)
      return;
    delete tree;
    tree = NULL;
  }