Create and calculate the arithmetic expression tree (for reference only)

Only to cope with the assignments assigned by the data structure teacher, the number in the formula can only be an integer: for example, input (6 + 3) * (4-2) * (41-1) output 720.00 # include # includeconstintN = 1000; intm [N], flag; doubleres; structNode {doubleval; charc; No

Only to cope with the assignments assigned by the data structure teacher, the number in the formula can only be an integer:

For example, input (6 + 3) * (4-2) * (41-1)

720.00 output

# Include
 
  
# Include
  
   
Const int N = 1000; int m [N], flag; double res; struct Node {double val; char c; Node * left; Node * right; Node () {left = right = NULL; c = 'A'-1 ;}}; Node * build (char * str, int l, int r) {int flag1 =-1, flag2 =-1, p = 0; Node * node = new Node; if (r-l = 1) {node-> val = m [str [l]; return node;} for (int I = l; I <r; I ++) {switch (str [I]) {case '(': p ++; break; case ')': p --; break; case '+': case '-': if (! P) flag1 = I; break; case '*': case '/': if (! P) flag2 = I; break;} if (flag1 <0) flag1 = flag2; if (flag1 <0) return build (str, l + 1, r-1 ); node-> left = build (str, l, flag1); node-> right = build (str, flag1 + 1, r); node-> c = str [flag1]; return node;} double preOrder (Node * root) {if (root = NULL) return 0; if (root-> c = 'A'-1) return root-> val; double a; switch (root-> c) {case '+': a = preOrder (root-> left) + preOrder (root-> right ); brea K; case '-': a = preOrder (root-> left)-preOrder (root-> right); break; case '*': a = preOrder (root-> left) * preOrder (root-> right); break; case '/': if (preOrder (root-> right )! = 0) a = preOrder (root-> left)/preOrder (root-> right); elseflag = 1; break;} return a;} int main () {Node * root; char str [N], s [N]; while (scanf ("% s", str )! = EOF) {memset (m, 0, sizeof (m); res = flag = 0; int len = strlen (str); int n = 0; char a = 'a'; if (str [0] = '-') s [n ++] = a ++; for (int I = 0; I <len; I ++) {if (str [I]> '9' | str [I] <'0 ') s [n ++] = str [I]; else {int sum = 0; while (str [I]> = '0' & str [I] <= '9') {sum = sum * 10 + str [I]-'0 '; I ++;} I --; m [a] = sum; s [n ++] = a ++;} s [n] = '\ 0 '; root = build (s, 0, n); res = preOrder (root); if (! Flag) printf ("expression result: %. 2lf \ n", res); elseprintf ("error! \ N ") ;}return 0 ;}