(PAT 1130) (二叉树加括号的表达式)

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本文介绍如何从语法树(binary)输出对应的中序表达式，重点在于如何正确添加括号反映运算符的优先级。通过中序遍历二叉树的方式，结合递归算法，实现表达式的正确构造。

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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.

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.Output Specification:

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))

解题思路:

中序遍历二叉树得到的就是一个表达式

这题的重点在怎样加括号上

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;

struct SDNode {
	string cdata;
	int lchild, rchild;
	bool isroot = true;
}SDTree[25];
int nroot = -1;
void midTraverse(int root) {
	if (root != nroot && (SDTree[root].lchild != -1 || SDTree[root].rchild != -1)) cout << "(";
	if (SDTree[root].lchild != -1) {
		midTraverse(SDTree[root].lchild);
	}
	cout << SDTree[root].cdata;
	if (SDTree[root].rchild != -1) {
		midTraverse(SDTree[root].rchild);
	}
	if (root != nroot && (SDTree[root].lchild != -1 || SDTree[root].rchild != -1)) cout << ")";
}

int main() {
	int N;
	cin >> N;
	for (int i = 1; i <= N; ++i) {
		string temdata;
		int lchi, rchi;
		cin >> temdata >> lchi >> rchi;
		SDTree[i].cdata = temdata;
		SDTree[i].lchild = lchi;
		SDTree[i].rchild = rchi;
		if (lchi != -1) SDTree[lchi].isroot = false;
		if (rchi != -1) SDTree[rchi].isroot = false;
	}
	//寻找根节点
	for (int i = 1; i <= N; ++i) {
		if (SDTree[i].isroot) {
			nroot = i;
			break;
		}
	}
	//遍历
	midTraverse(nroot);
	system("PAUSE");
	return 0;
}