怎么求二叉树的结点数，并返回结点数。下面的程序错在哪啊？？？求助啊！！！
#include "stdafx.h"
#include "stdlib.h"
typedef int ElemType;
struct BTreeNode{
    ElemType data;
    struct BTreeNode *left,*right;
};
  
int BTreeCount(struct BTreeNode* BT)
{   int n=0,m=0;
    if(BT=NULL)
      return (0);
    else
    {
       BTreeCount(BT->left );
         n++;
        
      
         BTreeCount(BT->right);
         m++;

      return (n+m);
      
      }
   
}

if(BT=NULL)这里写错了吧,应该是:
if(BT == NULL)

原来是这样！！！呵呵，谢谢

改完这个：if(BT=NULL)这里写错了吧,应该是:
if(BT == NULL)  也不对吧

int BTreeCount(struct BTreeNode* BT)
{
    if (!BT)           return 0;
    if (!BT->left)     return BTreeCount(BT->right) + 1;
    if (!BT->right)    return BTreeCount(BT->left)  + 1;
    return BTreeCount(BT->right) + BTreeCount(BT->left)  + 1;
}

int nodes(btnode *b)//求二叉树的节点个数
{
    int num1,num2;
    if(b==NULL)
        return 0;
    else if(b->lchild==NULL && b->rchild == NULL)
        return 1;
    else
    {
        num1 = nodes(b->lchild);
        num2 = nodes(b->rchild);
        return (num1+num2+1);
    }
}