数据结构实验三不会的
1、二叉树表达式求值
基于二叉树的表达式求值
描述
输入一个表达式(表达式中的数均为小于10的正整数),利用二叉树来表示该表达式,创建表达式树,然后利用二叉树的遍历操作求表达式的值。
输入
多组数据。每组数据一行,为一个表达式,表达式以‘=’结尾。当输入只有一个“=”时,输入结束。
输出
每组数据输出一行,为表达式的值。
输入样例 1
2*(2+5)= 1+2= =
输出样例 1
14 3
#include<iostream> #include<stack> #include<string> #include<cstring> #include<queue> using namespace std; typedef struct Node* BinTree; typedef BinTree BT; string s; queue<char> num; queue<char> op; struct Node { char Data; BT Left; BT Right; int ans; }; int fact(char c) { if (c >= '0' && c <= '9') return 1; else return 2; } BT createNode(char c) { BT p = new Node; p->Data = c; p->Left = p->Right = NULL; if (fact(c) == 1) p->ans = c - '0'; else p->ans = 0; return p; } BT createTree() { for (int i = 0; i < s.size() - 1; i++) { if (fact(s[i]) == 1) num.push(s[i]); else op.push(s[i]); } BT Head = NULL; int flag = 0; //标记有括号时的情况 int sflag = 0; //处理开始时为括号的情况 if (s[0] == '(') sflag = 1; while (!op.empty())//非空 { char option; option = op.front();//取第一个运算符 op.pop(); if (option != '(' && option != ')') { BT T = createNode(option); if (option == '+' || option == '-') { if (flag == 0) { if (Head == NULL) { T->Left = createNode(num.front()); num.pop(); T->Right = createNode(num.front()); num.pop();//取两个数,作为两个孩子 } else { T->Left = Head; T->Right = createNode(num.front()); num.pop(); } Head = T; } else { if (Head == NULL) { T->Left = createNode(num.front()); num.pop(); T->Right = createNode(num.front()); num.pop(); Head = T; } else { T->Left = Head->Right; Head->Right = T; T->Right = createNode(num.front()); num.pop(); } } } else if (option == '*' || option == '/') { if (flag == 0) { if (Head == NULL) { T->Left = createNode(num.front()); num.pop(); T->Right = createNode(num.front()); num.pop(); Head = T; } else { if (sflag == 1 || Head->Data == '*' || Head->Data == '/') { T->Left = Head; Head = T; T->Right = createNode(num.front()); num.pop(); sflag = 0; } else { T->Left = Head->Right; Head->Right = T; T->Right = createNode(num.front()); num.pop(); } } } if (flag == 1) { if (Head == NULL) { T->Left = createNode(num.front()); num.pop(); T->Right = createNode(num.front()); num.pop(); Head = T; } else { T->Left = Head->Right; Head->Right = T; T->Right = createNode(num.front()); num.pop(); } } } } else if (option == '(') { flag = 1; } else if (option == ')') { flag = 0; } } return Head; } void solve(BT L) { if (L) { solve(L->Left); solve(L->Right); char option = L->Data; if (option == '+') L->ans = L->Left->ans + L->Right->ans; if (option == '-') L->ans = L->Left->ans - L->Right->ans; if (option == '*') L->ans = L->Left->ans * L->Right->ans; if (option == '/') L->ans = L->Left->ans / L->Right->ans; } } int main() { while (cin >> s && s[0] != '=') { BT H = createTree(); solve(H); cout << H->ans << endl; } return 0; }