【PAT】【Advanced Level】1130. Infix Expression (25)

1130. Infix Expression (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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))

原题链接：

https://www.patest.cn/contests/pat-a-practise/1130

思路：

树的中序遍历

根节点、叶节点不加括号，其他节点加括号

注意只有一项的情况

CODE：

#include
#include
#include
#include
#define N 22
using namespace std;
typedef struct S
{
	string str;
	int ls;
	int rs;	
};
S T[N];
bool flag[N];
vector res;
void dfs(int n)
{
	if (T[n].ls==-1 && T[n].rs==-1)
	{
		res.push_back(T[n].str);
		return ;
	}
	res.push_back("(");
	if (T[n].ls!=-1)
		dfs(T[n].ls);
	res.push_back(T[n].str);
	if (T[n].rs!=-1)
		dfs(T[n].rs);
	res.push_back(")");
}
int main()
{
	memset(flag,0,sizeof(flag));
	int n;
	cin>>n;
	for (int i=1;i<=n;i++)
	{
		S ne;
		cin>>ne.str>>ne.ls>>ne.rs;
		if (ne.ls!=-1)
		{
			flag[ne.ls]=1;
		}
		if (ne.rs!=-1)
		{
			flag[ne.rs]=1;
		}
		T[i]=ne;
	}
	int head;
	for (int i=1;i<=n;i++)
	{
		if (flag[i]==0)
		{
			head=i;
			break;
		}
	}
	dfs(head);
	for (int i=1;i