二叉树之叶子节点个数

二叉树中求叶子节点个数:
当二叉树不为空,而且左子树和右字数 不为空时:
#include 
#include 
typedef char ElemType;
typedef struct BiTNode{
ElemType data;
struct BiTNode* Lchild;
struct BiTNode* Rchild;
}BiTNode,*BiTree;
//构造二叉树
void createbt(BiTree &T)
{
    char ch;
    scanf("%c",&ch);
     if(ch=='#')
		 T=NULL;
     else
	 {
           T=(BiTree)malloc(sizeof(BiTNode));
           T->data=ch;
           createbt(T->Lchild);
           createbt(T->Rchild);
	 }
}
//统计二叉树中叶子节点的个数
void Numnode(BiTree T,int &num)
{
   if(T)
   {
     if((!T->Lchild)&&(!T->Rchild))
       num++;
     Numnode(T->Lchild,num);
     Numnode(T->Rchild,num);
   }
}
int main()
{
   int count=0;
    BiTree T;
    printf("create a tree such as ABC##DE#G##F###
");
    createbt(T);
    printf("统计二叉树中叶子节点的个数
");
    Numnode(T,count);
    printf("%d
",count);
	return 0;
}
原文地址:https://www.cnblogs.com/NYNU-ACM/p/4236864.html