求二叉树的结点个数，如果根节点为空，则返回 0。
#include<iostream>#include <malloc.h>
using namespace std;
static int D = 0;//记录具有两个子女结点的结点个数
//定义二叉树结构
typedef struct Bintreenode {
char data;
struct Bintreenode *right;
struct Bintreenode *left;
};
//先序创建二叉树
void CreatBiTree(Bintreenode *&T){
char ch;
cin >> ch;
if (ch == '#') {
T = NULL;
}
else {
T = (Bintreenode*)malloc(sizeof(Bintreenode));
T->data = ch;
CreatBiTree(T->left);//递归创建左子树
CreatBiTree(T->right);//递归创建右子树
}
}
//先序遍历二叉树
void Preorder(Bintreenode *&T) {
if (T) {
cout << T->data;
Preorder(T->left);
Preorder(T->right);
}
else
cout << "";
}
//二叉树结点个数
int CountNode(Bintreenode *&root) {
int count = 0;
if (root == NULL)
return 0;
int m=CountNode(root->left);
int n=CountNode(root->right);
if (root->left != NULL && root->right != NULL)//判断该结点是否有两个子女结点
D++;
count=m+n+1;
return count;
}
int main()
{
cout << "创建一个二叉树：" << endl;
Bintreenode *T;
CreatBiTree(T);
cout << "先序遍历：" << endl;
Preorder(T);
cout << endl;
cout << "二叉树的结点个数" << endl;
cout << CountNode(T)<<endl;
cout << "二叉树的具有两个子女结点的结点个数" << endl;
cout << D << endl;
return 0;
}
运行结果：
