二叉树度数为0，1，2，所对应的结点个数

二叉树度数为0，1，2，所对应的结点个数

#include<stdio.h>
#include<stdlib.h>
#include<malloc.h>
typedef struct node{
	char data;
	struct node *Lchild;
	struct node *rchild;
}BiTnode;
//建立二叉树
BiTnode *Create_Tree(){
	char ch;
	BiTnode *root;
	scanf("%c",&ch);
	if(ch=='#') root=NULL;
	else{
		root=(BiTnode *)malloc(sizeof(BiTnode));
		root->data=ch;
		root->Lchild=Create_Tree();
		root->rchild=Create_Tree();
	}
	return root;
}
int count1=0,count2=0,count3=0;
//查找各度数结点所对应的个数
int CountNode(BiTnode *root){
	if(root){
		CountNode(root->Lchild);
		CountNode(root->rchild);
		if(root->Lchild==NULL&&root->rchild==NULL){
			count1++;
		}else if(root->Lchild!=NULL&&root->rchild==NULL){
			count3++;
		}else{
			count2++;
		}
	}
	return count2;
}
int main(){
	BiTnode *T;
	T=Create_Tree();
	int n=CountNode(T);
	//printf("度为0的结点个数为：");
	//printf("%d\n",count1);
	printf("度为1的结点个数为：");
	printf("%d\n",n);
	//printf("度为2的结点个数为：");
	//printf("%d\n",count3);
	return 0;
}