SDUT2484 算术表达式的转换(表达式树)

#include <stdio.h>
#include <string.h>

#define maxn 1000

int lch[maxn], rch[maxn], nc = 0;
char op[maxn];

int build_tree(char *s, int x, int y) {
int i, c1 = -1, c2 = -1, p = 0;
int u;

if(y-x == 1) {
u = ++nc;
lch[u] = rch[u] = 0; op[u] = s[x];
return u;
}

for(i=x; i<y; i++) {
switch(s[i]) {
case '(': p++; break;
case ')': p--; break;
case '+': case '-': if(!p) c1 = i; break;
case '*': case '/': if(!p) c2 = i; break;
}
}

if(c1 < 0) c1 = c2;
if(c1 < 0) return build_tree(s, x+1, y-1);

u = ++nc;
lch[u] = build_tree(s, x, c1);
rch[u] = build_tree(s, c1+1, y);
op[u] = s[c1];

return u;
}

void pre_print(int u) {
if(u != 0) {
printf("%c", op[u]);
pre_print(lch[u]);
pre_print(rch[u]);
}
}

void in_print(int u) {
if(u != 0) {
in_print(lch[u]);
printf("%c", op[u]);
in_print(rch[u]);
}
}

void las_print(int u) {
if(u != 0) {
las_print(lch[u]);
las_print(rch[u]);
printf("%c", op[u]);
}
}

int main() {
char s[5000];
while(scanf("%s", s) == 1) {

build_tree(s, 0, strlen(s)-1);

pre_print(1);
putchar('\n');

in_print(1);
putchar('\n');

las_print(1);
putchar('\n');
}

return 0;
}