Description
Given a preorder sequence, create a corresponding binary tree according to the sequence, and output the number of nodes with degrees 0, 1 and 2 of the binary tree.
Input
One line, the binary tree traverses the sequence in preorder, and the null pointer is occupied by the character ^
Output
One line, three integers represent the number of nodes
with degrees
Sample Input
ABD ^^^ CE ^^ F ^^
Sample Output
3 1 2
Recursively traverse to find! ! ! ! ! !
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct node{
char ch;
struct node *Lchild;
struct node *Rchild;
}BiTNode,*BiTree;
void Q_CreatTree(BiTree *T);
int NodeNumber_1(BiTree T);
int NodeNumber_2(BiTree T);
int NodeNumber_0(BiTree T);
int main(void)
{
BiTree T;
Q_CreatTree(&T);
getchar();
printf("%d ",NodeNumber_0(T));//Node with degree 0
printf("%d ",NodeNumber_1(T));//Node with degree 1
printf("%d",NodeNumber_2(T));//Node with degree 2
}
void Q_CreatTree(BiTree *T)//Preorder traversal to build a binary tree
{
char str;
str=getchar();
if(str=='^')
{
*T=NULL;
}
else
{
(*T)=(BiTree)malloc(sizeof(BiTNode));
(*T)->ch=str;
Q_CreatTree( &( (*T)->Lchild ) );
Q_CreatTree( &( (*T)->Rchild ) );
}
}
int NodeNumber_0(BiTree T)
{
int i=0;
if(T)
{
if(T->Lchild==NULL&&T->Rchild==NULL)
{
i=1;
}
else
{
i=NodeNumber_0( T->Lchild ) + NodeNumber_0( T->Rchild );
}
}
return i;
}
int NodeNumber_1(BiTree T)
{
int i=0;
if(T)
{
if( (T->Lchild==NULL&&T->Rchild!=NULL) ||(T->Lchild!=NULL&&T->Rchild==NULL))
{
i=1+NodeNumber_1(T->Lchild)+NodeNumber_1(T->Rchild);
}
else
{
i=NodeNumber_1(T->Lchild)+NodeNumber_1(T->Rchild);
}
}
return i;
}
int NodeNumber_2(BiTree T)
{
int i=0;
if(T)
{
if((T->Lchild!=NULL)&&(T->Rchild!=NULL))
{
i=1+NodeNumber_2(T->Lchild) + NodeNumber_2(T->Rchild);
}
else
{
i= NodeNumber_2(T->Lchild) + NodeNumber_2(T->Rchild);
}
}
return i;
}