【数据结构】（二叉树）递归求解双分支结点个数

算法思想： 求二叉树双分支结点个数就是当一个结点左右孩子非空
时该结点为双分支结点

---

//二叉树结构体定义
typedef struct BiTree{
	char data;
	BiTree *lchild;
	BiTree *rchild;
	
}BiTree;

int DsonNodes(BiTree *T){
	if(!T){
		return 0;
	}
	if(T->rchild!=NULL && T->lchild!=NULL) //若结点的左右孩子存在
		return 1+DsonNodes(T->rchild)+DsonNodes(T->lchild);
	else
		return DsonNodes(T->rchild)+DsonNodes(T->lchild);

}

测试数据：

	char pre[13]={'#','A','B','D','F','H','J','M','N','C','E','K','G'};
	char in[13]={'#','D','B','J','H','M','N','F','A','E','K','C','G'};

	char pre1[10]={'#','A','B','C','D','E','F','G','H','I'};
	char in1[10]={'#','B','C','A','E','D','G','H','F','I'};

	char pre2[6]={'#','A','B','D','E','C'};
	char in2[6]={'#','D','B','E','A','C'};

	char pre3[5]={'#','A','B','D','C'};
	char in3[5]={'#','D','B','A','C'};

	cout<<"The number of double-branch nodes is: "<<DsonNodes(PreInCreate(pre,in,1,12,1,12))<<endl;
	cout<<"The number of double-branch nodes is: "<<DsonNodes(PreInCreate(pre3,in3,1,4,1,4))<<endl;
	cout<<"The number of double-branch nodes is: "<<DsonNodes(PreInCreate(pre1,in1,1,9,1,9))<<endl;

结果：