PAT - 甲级 - 1130. Infix Expression (25) （中序遍历）

题目描述：

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))

题目思路：

给定中缀表达式二叉树，输出中缀表达式。需要注意的是括号（parentheses）的输出，二叉树的根节点和叶子结点不需要输出括号。在程序中加以限制即可。

题目代码：

#include
#include
#include
#include
#define MAXN 21
using namespace std;
int n, Root;
int vis[MAXN];

struct Node{
	string s;
	int l, r;
}node[MAXN];


void dfs(int root)
{	//border 
	if(root==-1) return ;
	//判断是否是根节点或者叶子结点 
	if(root!=Root&&(node[root].l!=-1||node[root].r!=-1)) cout<<"(";
	//中序遍历 
	dfs(node[root].l);
	cout<>node[i].s>>node[i].l>>node[i].r;
		vis[node[i].l] = vis[node[i].r] = 1;
	}
	//找到根节点 
	Root = 1;
	while(vis[Root])Root++;
	
	dfs(Root);
	
	return 0;
}