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！！！

qinyao 发布于 2013-05-09 19:29， 1401 次点击

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);

      }

}

5 回复

笑傲2013-05-09 20:24

if(BT=NULL)这里写错了吧,应该是:
if(BT == NULL)

qinyao2013-05-09 21:25

原来是这样！！！呵呵，谢谢

走召2013-05-09 23:14

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

azzbcc2013-05-09 23:22

程序代码：

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;
}

计科11022013-05-10 22:43

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);
    }
}