Codeforces Round #287 (Div. 2) C. Guess Your Way Out! 二叉树遍历

思路：由样例可知，若exit和实际command在当前的父节点两边，则父节点一侧的子树节点必须全部遍历一遍，才会再次返回到父节点，从而转向正确的方向。

故先求出exit的正确路径，然后与command相比，

①若相同，则只需沿着command走，ans++

②若不相同，则ans += 父节点左侧子树的节点数

代码如下：

#include <cstdio>
using namespace std;
#define N 51
#define LEFT 0
#define RIGHT 1
long long fac
, sum
;
int dir
;

int main(){
int h;
long long n;
scanf("%d %I64d", &h, &n);
fac[0] = 1, sum[0] = 1;
for(int i = 1; i <= h; ++i)
fac[i] = fac[i - 1] * 2, sum[i] = sum[i - 1] + fac[i];
int level = h;
long long ans = 0, cur = 0;
int cmd = LEFT;
for(int i = h; i > 0; --i){
if(n > cur + fac[i - 1])
dir[i] = RIGHT, cur += fac[i - 1];
}
while(level){
if(cmd == dir[level]){
++ans, --level;
}
else
ans += sum[level - 1];
cmd = 1 - cmd;
}
printf("%I64d\n", ans);
return 0;
}