PAT 甲级 1130 Infix Expression-优快云博客

本文介绍了一种从语法树（二叉树形式）转换为对应的带括号中缀表达式的算法实现，该算法通过深度优先搜索（DFS）递归地遍历树结构，确保运算符的优先级正确反映在最终的表达式中。

https://pintia.cn/problem-sets/994805342720868352/problems/994805347921805312

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 ( $\leq 20) which is the total number of nodes in the syntax tree. Then N lines follow, each gives the information of a node (the$

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 $-. The figures 1 and 2 correspond to the samples 1 and 2, respectively.$


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))

代码：

#include <bits/stdc++.h>
using namespace std;

int N;
int vis[30];

struct Node{
    string val;
    int L, R;
}node[30];

string dfs(int st) {
    if(node[st].L == -1 && node[st].R == -1) return node[st].val;
    if(node[st].L == -1 && node[st].R != -1) return "(" + node[st].val + dfs(node[st].R) + ")";
    if(node[st].L != -1 && node[st].R != -1) return "(" + dfs(node[st].L) + node[st].val + dfs(node[st].R) + ")";
}

int main() {
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++) {
        cin >> node[i].val >> node[i].L >> node[i].R;
        if(node[i].L != -1) vis[node[i].L] = 1;
        if(node[i].R != -1) vis[node[i].R] = 1;
    }

    int root = 1;
    while(vis[root] == 1) root ++;
    string ans = dfs(root);
    if(ans[0] == '(') ans = ans.substr(1, ans.size() - 2);
    cout << ans;
    return 0;
}

　　dfs 递归最后去掉最外面的括号

可能最近是自闭 girl 了希望有好运气叭

转载于:https://www.cnblogs.com/zlrrrr/p/10417606.html