PAT 1162 Postfix Expression

1162 Postfix Expression (25 分)

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)%))+)

 思路：

对节点的孩子情况进行分类：

1.两个孩子的进行左-右-根的遍历

2.一个孩子的进行孩子-根的遍历

3.没孩子直接输出

#include<bits/stdc++.h>
using namespace std;
#define ll long long
string c[25];
int lson[25];
int rson[25];
int b[25];
void dfs(int x)
{
	printf("(");
	if (lson[x] == -1&&rson[x]==-1)
	{
		cout<<c[x];
	}
	else if(lson[x]==-1)
	{
		cout<<c[x];
		dfs(rson[x]);
	}
	else if (rson[x] == -1)
	{
		cout<<c[x];
		dfs(lson[x]);
	}
	else
	{
		dfs(lson[x]);
		dfs(rson[x]);
		cout<<c[x];
	}
	printf(")");
}
int main()
{
	int n; cin >> n;
	for (int i = 1; i <= n; ++i)
	{
		cin >> c[i];
		cin >> lson[i] >> rson[i];
		b[lson[i]] = 1;
		b[rson[i]] = 1;
	}
	int root = 0;
	for (int i = 1; i <= n; ++i)
	{
		if (b[i] == 0)root = i;
	}
	dfs(root);
}