DSOJ Specular reflection of a tree（树的镜面映射）

最新推荐文章于 2023-05-24 23:04:17 发布

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最新推荐文章于 2023-05-24 23:04:17 发布

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分类专栏： Data Structure 文章标签： DSOJ 树镜面映射

本文链接：https://blog.youkuaiyun.com/xhf0374/article/details/50360491

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本文详细探讨了DSOJ中的一道题目，涉及树的镜面映射问题。通过分析和解题过程，揭示了树结构在镜面映射算法中的应用。

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题目链接

#include<stdio.h>			//Specular reflection of a tree
#include<string.h>
#include<stdlib.h>
#define MAX 100				//树的最大结点数

typedef struct node
{
	char info;
	int level;					//记录结点所在的层数，便于控制输出树的镜面映射的广度优先遍历
	struct node *left, *right;
}BT;

void BFS(BT *T)			//BFS of the Binary Tree
{
	BT *q[MAX], *p, *pa;
	char stack[MAX];
	int front = 0, rear = 0, top = 0;
	if (!T) return;
	q[rear++] = T;
	T->level = 1;
	while (front < rear)		//将镜像映射之前的树的深度优先遍历结果存储在队列中，并记录每一个结点的层数
	{
		pa = q[front++];
//		printf("%c %d\n", pa->info,pa->level);
		p = pa->left;
		while (p)
		{
			p->level = pa->level + 1;
			q[rear++] = p;
			p = p->right;
		}
	}
	front = 0;
	while (front < rear)		//利用栈，将同一层的结点全部入栈在一次出栈，从而达到镜像映射的目的
	{
		stack[top++] = q[front]->info;
		while (front < rear - 1 && q[front]->level == q[front + 1]->level)
		{
			stack[top++] = q[front+1]->info;
			front++;
		}
		while (top > 0)
			printf("%c ", stack[--top]);
		if (front < rear - 1 && q[front]->level != q[front + 1]->level)
			front++;
		else
			break;
	}
	printf("\n");
}

BT *create()		//Create the Binary Tree
{
	BT *T;
	int flag;
	char c;
	T = (BT *)malloc(sizeof(BT));
	c = getchar();				//Receive Enter and blank
	c = getchar();
	scanf("%d", &flag);
	if (c == '$') return NULL;	//Virtual node
	if (c != '$'&&flag == 1)	//Leaf node
	{
		T->info = c;
		T->left = NULL;
		T->right = NULL;
		return T;
	}
	T->info = c;				//非终端结点
	T->left = create();
	T->right = create();
	return T;
}

int main()
{
	BT *T;
	int N;			
	scanf("%d", &N);			//N 并没有什么卵用
	T = create();
	BFS(T);
	return 0;
}