PAT 甲级 1130 Infix Expression

1130 Infix Expression （25 point(s)）

Given a syntax tree (binary), you are supposed to output the corresponding infix expression, with parentheses reflecting the precedences of the operators.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 20) which is the total number of nodes in the syntax tree. Then N lines follow, each gives the information of a node (the i-th line corresponds to the i-th node) in the format:

data left_child right_child

where data is a string of no more than 10 characters, left_child and right_child are the indices of this node's left and right children, respectively. The nodes are indexed from 1 to N. The NULL link is represented by −1. The figures 1 and 2 correspond to the samples 1 and 2, respectively.

Figure 1	Figure 2

Output Specification:

For each case, print in a line the infix expression, with parentheses reflecting the precedences of the operators. Note that there must be no extra parentheses for the final expression, as is shown by the samples. There must be no space between any symbols.

Sample Input 1:

8
* 8 7
a -1 -1
* 4 1
+ 2 5
b -1 -1
d -1 -1
- -1 6
c -1 -1

Sample Output 1:

(a+b)*(c*(-d))

Sample Input 2:

8
2.35 -1 -1
* 6 1
- -1 4
% 7 8
+ 2 3
a -1 -1
str -1 -1
871 -1 -1

Sample Output 2:

(a*2.35)+(-(str%871))

经验总结：

408计算机基础综合上面的DS算法题，利用中序遍历，第一层以及叶结点不需要输出括号，其他的均需要输出括号，难度不大，就不多说啦~

AC代码

#include <cstdio>
#include <set>
using namespace std;
const int maxn=25;
int n;
bool flag[maxn]={false};
struct node
{
	int lchild,rchild;
	char data[15];
}Node[maxn];
void in_order(int root,int depth)
{
	if(root==-1)
		return ;
	bool flag=true;
	if(Node[root].lchild==-1&&Node[root].rchild==-1)
		flag=false;
	if(flag&&depth!=1)
		printf("(");
	in_order(Node[root].lchild,depth+1);
	printf("%s",Node[root].data);
	in_order(Node[root].rchild,depth+1);
	if(flag&&depth!=1)
		printf(")");
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%s%d%d",Node[i].data,&Node[i].lchild,&Node[i].rchild);
		if(Node[i].lchild!=-1)
			flag[Node[i].lchild]=true;
		if(Node[i].rchild!=-1)
			flag[Node[i].rchild]=true;
	}
	int root;
	for(root=1;root<=n;++root)
	{
		if(flag[root]==false)
			break;
	}
	in_order(root,1);
	return 0;
}