机试指南 3.3二叉树 含练习题

机试指南 3.3二叉树

例3.4二叉树遍历

#include<stdio.h>
#include<string.h>

struct node{
	node *lchild;
	node *rchild;
	char c;
}tree[50];
int loc;

char str1[30];
char str2[30];

node *create();
void postOrder(node *t);
node *build(int s1,int e1,int s2,int e2);

int main()
{
	while(scanf("%s",str1)!=EOF)
	{
		scanf("%s",str2);
		int l1=strlen(str1);
		int l2=strlen(str2);
		loc=0;
		node *T=build(0,l1-1,0,l2-1);
		postOrder(T);
		printf("\n");
	}
	return 0;
}

node *create()
{
	tree[loc].lchild=NULL;
	tree[loc].rchild=NULL;
	return &tree[loc++];
}

void postOrder(node *t)
{
	if(t->lchild!=NULL)
		postOrder(t->lchild);
	if(t->rchild!=NULL)
		postOrder(t->rchild);
	printf("%c",t->c);
}

node *build(int s1,int e1,int s2,int e2)
{
	node *ret=create();
	ret->c=str1[s1];
	
	int idx;
	for(int i=s2;i<=e2;i++)
	{
		if(str2[i]==ret->c)
		{
			idx=i;
			break;
		}
	}
	
	if(idx!=s2)
		ret->lchild=build(s1+1,s1+(idx-s2),s2,idx-1);
	
	if(idx!=e2)
		ret->rchild=build(s1+(idx-s2)+1,e1,idx+1,e2); 
	
	return ret;
}

练1：二叉树KY85

#include<stdio.h>

int countx(int t,int n)
{
	if(t<=n)
	{
		if( (t*2<=n) && ((t*2+1)<=n) )
			return 1+countx(t*2,n)+countx(t*2+1,n);
		else if( (t*2<=n) && ((t*2+1)>n) )
			return 1+countx(t*2,n);
		else if(t*2>n)
			return 1;
	}
	else
		return 0; 
}
int main()
{
	int m,n;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		if(m==0&&n==0)
			break;
		int result;
		result=countx(m,n);
		printf("%d\n",result);
	}
	return 0;
}

练1：二叉树KY47

#include<stdio.h>
#include<math.h>

int degree(int t)
{
	int d=1;
	while(true)
	{
		if(pow(2,d-1)>t)
			break;
		d++;
	}
	return --d;
}
int main()
{
	int x,y;
	while(scanf("%d%d",&x,&y)!=EOF)
	{
		if(x==y)
		{	
			printf("%d\n",x);
			continue;
			}	
		if(degree(x)>degree(y))
		{
			int z=x;
			x=y;
			y=z;
		}
		
		if(degree(x)<degree(y))
		{
			int a=degree(y)-degree(x);
			for(int i=0;i<a;i++)
				y=y/2;
		}
		
		while(true)
		{
			if(x==y)
				break;
			x=x/2;
			y=y/2;
		}
		printf("%d\n",x);
	}
	return 0;
}

练2：树查找

#include<stdio.h>
#include<math.h>

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int buf[1000];
		for(int i=1;i<=n;i++)
			scanf("%d",&buf[i]);
		int count;
		scanf("%d",&count);
		if(pow(2,count-1)<=n)
		{
			int x=pow(2,count-1);
			printf("%d",buf[x]);
			while(true)
			{
				x++;
				if(x>=pow(2,count))
					break;
				printf(" %d",buf[x]);
			}
			printf("\n");
		}
		else
			printf("EMPTY\n");
	}
	return 0;
}