Swust OJ975: 统计利用先序遍历创建的二叉树的度为2的结点个数

void DegreeTwo(Tree *&tree)
{
    if(tree!=NULL)
    {
        if(tree->lchild!=NULL&&tree->rchild!=NULL)
//左右孩子均存在，该结点为二叉树度为2的结点
            count++;
        if(tree->lchild!=NULL)
            DegreeTwo(tree->lchild);
//当不满足左右孩子均存在，单独判断左孩子是否有它的左右孩子
        if(tree->rchild!=NULL)
            DegreeTwo(tree->rchild);
//右孩子同上
    }
}

#include<iostream>
#include<malloc.h>
using namespace std;
int count=0;
typedef struct node
{
    char data;
    struct node *lchild,*rchild;
}Tree;
void CreateTree(Tree *&tree)
{
    char ch;
    cin>>ch;
    if(ch==‘#‘)
        tree=NULL;
    else
    {
        tree=(Tree*)malloc(sizeof(Tree));
        tree->data=ch;
        CreateTree(tree->lchild);
        CreateTree(tree->rchild);
    }
}
void DegreeTwo(Tree *&tree)
{
    if(tree!=NULL)
    {
        if(tree->lchild!=NULL&&tree->rchild!=NULL)
            count++;
        if(tree->lchild!=NULL)
            DegreeTwo(tree->lchild);
        if(tree->rchild!=NULL)
            DegreeTwo(tree->rchild);
    }
}
int main()
{
    Tree *tree;
    CreateTree(tree);
    DegreeTwo(tree);
    cout<<count;
    return 0; 
}