二叉树的前中后序遍历以及表达式树

if (ch == '-') return a - b;
if (ch == '*') return a * b;
if (ch == '%') return (int)((int)a % (int)b);
if (ch == '/') {
if (b != 0) return a / b;
error = true; return 0;
error = false;
int len = s.length (), i = 0;
while (i < len) {
if (s[i] == ' ' || s[i] == '=') {i++; continue;}
else if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/' || s[i] == '%')
void print(btree p) {
if (! (p -> lch) && ! (p -> rch)) {
cout << fixed << setprecision (1) << p -> data << " ";
//
cout << p -> data << " ";
}
else cout << p -> op << " ";
while (i >= 0) {
if (s[i] == ' ' || s[i] == '=') {
i--; continue;
}
else if (s[i] == ')') {
op.push (s[i--]);
}
else if (s[i] == '(') {
while (op.top () != '#' && op.top () != ')') {
举个栗子：表达式：1 + 2 * (3 - 4) - 5 / 6，那么它的前缀表达式就是：- + 1 * 2 - 3 4 / 5 6， 那么可以建立二叉树，如图：
最后，附上代码
/************************************************
* Author
:你们亲爱的室友
{
double a = num.top (); num.pop ();
double b = num.top (); num.pop ();
num.push (cal (b, a, s[i])); i++;
}
else {
double x = 0;
while (is_digit (s[i])) {
x = x * 10 + s[i] - '0'; i++;
}
//中序遍历
void post_order(btree T) { //后序遍历 if (T != NULL) { post_order (T -> lch); post_order (T -> rch); print (T); }
}
double cal_tree(btree &T) { if (T != NULL) { if (! (T -> lch) && ! (T -> rch)) {
ret += op.top (); ret += ' ';
op.pop ();
}
op.push (s[i--]);
}
else {
while (is_digit (s[i]) || s[i] == '.') {
ret += s[i--];
}
ret += ' ';
}
}
while (op.top () != '#') {
T -> data = get_num (s); T -> lch = T -> rch = NULL; i++; return ;
} else {
T -> op = s[i]; i += 2; creat (T -> lch, s); creat (T -> rch, s); } }
void pre_order(btree T){
二叉树的前中后序遍历以及表达式树
昨天数据结构课布置了上机实验，要求递归方式建立表达式二叉树，输出树的前中后序遍历的结 果，并计算表达式的值。网上其他人的做法无非就是先求出后缀表达式，然后后序遍历的方式+栈建 立二叉树，可是本题的要求是递归方式，所以我的方法就是求出前缀表达式，用前序遍历的方法可以 递归建立二叉树，最后用后序遍历的方式求解表达式树。
} }
return T -> data; } else return E.cal (cal_tree (T -> lch), cal_tree (T -> rch), T -> op);
int main(void) { ios::sync_with_stdio (false); int T; cin >> T; string str; getline (cin, str); while (T--) { getline (cin, str); string pre = E.get_prefix (str), post = E.get_postfix (str); cout << "前缀表达式：" << pre << endl; cout << "后缀表达式：" << post << endl;
if (s[i] == ' ' || s[i] == '=') {
i++; continue;
}
else if (s[i] == '(') {
op.push (s[i++]);
}
else if (s[i] == ')') {
while (op.top () != '(') {
ret += op.top (); ret += ' ';
ret += op.top (); ret += ' ';
op.pop ();
}
op.push (s[i++]);
}
else {
while (is_digit (s[i]) || s[i] == '.') {
ret += s[i++];
}
ret += ' ';
}
}
while (op.top () != '#') {
* Created Time :2015/10/29 星期四 16:12:53
* File Name :BT.cpp
************************************************/
#include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> #include <iomanip> using namespace std;
default: return 0;
}
}
string get_prefix(string s) {
//中缀表达式转变前缀表达式
while (!op.empty ()) op.pop ();
op.push ('#');
string ret = "";
int len = s.length (), i = len - 1;
ret += op.top (); ret += ' ';
op.pop ();
}
ret += '=';
return ret;
}
double solve(string str) {
//计算后缀表达式
string s = get_postfix (str);
while (!num.empty ()) num.pop ();
}
void creat(btree &T, string s) { T = new node;
while (i < s.length () && (s[i] == ' ' || s[i] == '.')) i++; if (i >= s.length ()) {
T -> lch = T -> rch = NULL; return ; } if (is_digit (s[i])) {
//前序遍历
if (T != NULL) {
print (T);
pre_order (T -> lch);
pre_order (T -> rch);
}
}
void in_order(btree T) { if (T != NULL) { in_order (T -> lch); print (T); in_order (T -> rch); }
表达式求值，逆波兰式(后缀表达式)算法 输入(可以有空格，支持小数，实现'+-/*%')： ((1+2)*5+1)/4= 注意：取模一定是要整型，实现版本数字全是 double，强制类型转换可能倒置错误 转换为后缀表达式： 得到：1 2 + 5 * 1 + 4 / = 计算后缀表达式：得到：4.00 */ bool is_digit(char ch) { return '0' <= ch && ch <= '9'; }
op.pop ();
}
op.pop (); i++;
}
else if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/' || s[i] == '%')
{
while (prior (op.top ()) >= prior (s[i])) {
}
if (s[i] == '.') {
double k = 10.0, y = 0;
i++;
while (is_digit (s[i])) {
y += ((s[i] - '0') / k);
i++; k *= 10;
} x += y; } num.push (x); } } return num.top (); } }E; typedef struct BT { char op; double data; BT *lch, *rch; }node, *btree;
ret += op.top (); ret += ' ';
op.pop ();
}
op.pop (); i--;
}
else if (s[i] == '+' || s[i] == '-' || s[i] == '*Biblioteka Baidu || s[i] == '/' || s[i] == '%')
{
while (prior (op.top ()) > prior (s[i])) {
}
return 0;
}
string get_postfix(string s) {
//中缀表达式转变后缀表达式
while (!op.empty ()) op.pop ();
op.push ('#');
string ret = "";
int len = s.length (), i = 0;
while (i < len) {
ret += op.top (); ret += ' ';
op.pop ();
}
reverse (ret.begin (), ret.end ());
return ret;
}
double cal(double a, double b, char ch) {
if (ch == '+') return a + b;
#define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = acos (-1.0); /*
struct Exp { stack<char> op; stack<double> num; bool error;
int prior(char ch) {
//运算符的优先级
switch (ch) {
case '+':
case '-': return 1;
case '*':
case '%':
case '/': return 2;
}
int i;
double get_num(string s) { double x = 0; while (is_digit (s[i])) { x = x * 10 + s[i] - '0'; i++; } if (s[i] == '.') { double k = 10.0, y = 0; i++; while (is_digit (s[i])) { y += ((s[i] - '0') / k); i++; k *= 10; } x += y; } return x;