Given a syntax tree (binary), you are supposed to output the corresponding postfix 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 postfix expression, with parentheses reflecting the precedences of the operators.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)%))+)
#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
#define INF 0x3f3f3f3f
struct node
{
string val;
int l,r;
}p[5000];
void dfs(int x)
{
if(p[x].l!=-1&&p[x].r!=-1)
{
printf("(");
if(p[x].l!=-1)dfs(p[x].l);
if(p[x].r!=-1)dfs(p[x].r);
cout<<p[x].val;
printf(")");
}
else
{
printf("(");
cout<<p[x].val;
if(p[x].l!=-1)dfs(p[x].l);
if(p[x].r!=-1)dfs(p[x].r);
printf(")");
}
}
int main()
{
int n,i,j,root;
int vis[5000]={0};
scanf("%d",&n);
for(i=1;i<=n;i++)
{
cin>>p[i].val>>p[i].l>>p[i].r;
if(p[i].l!=-1)
vis[p[i].l]=1;
if(p[i].r!=-1)
vis[p[i].r]=1;
}
for(i=1;i<=n;i++)
{
if(vis[i]==0)
{
root=i;
break;
}
}
dfs(root);
}