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))
#include<iostream>
#include<vector>
#include<string>
using namespace std;
struct node{
string data;
int l,r;
}t[101];
string dfs(int i){
if(t[i].l==-1&&t[i].r==-1){
return t[i].data;
}
if(t[i].l!=-1&&t[i].r==-1){
return "("+dfs(t[i].l)+t[i].data+")";
}
if(t[i].l==-1&&t[i].r!=-1){
return "("+t[i].data+dfs(t[i].r)+")";
}
if(t[i].l!=-1&&t[i].r!=-1){
return "("+dfs(t[i].l)+t[i].data+dfs(t[i].r)+")";
}
}
int main(){
int n;
vector<int> book(101,0);
cin >> n;
for(int i = 1;i <= n;i++){
cin >> t[i].data >> t[i].l >> t[i].r;
if(t[i].l != -1) book[t[i].l] = 1;
if(t[i].r != -1) book[t[i].r] = 1;
}
int j = 1;
while(book[j]==1){
j++;
}
string tree = dfs(j);
if(tree[0]=='('){
tree = tree.substr(1,tree.size()-2);
}
cout << tree;
return 0;
}
语法树到中缀表达式转换
本文介绍了一种从二叉语法树转换为带有括号反映运算符优先级的中缀表达式的算法。通过递归遍历节点,正确地处理了单子树和双子树情况,确保了最终表达式的正确性和简洁性。


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