Description
给你一棵有 $n$ 个点的搜索二叉树的层序遍历序，问你它是不是一棵正则二叉树。单个测试点内共有 $T$ 组数据。
Solution
拿个队列存一下当前没确定是否放儿子的节点，然后直接模拟一下即可，在过程中存一下一个节点的值域限制即可。
正确性可以证明。
Code
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| #include <cstdio>
#define inf 999999999
#define maxN 100010
int a[maxN][3];
int f[maxN];
int read ()
{
int x = 0;
char c = getchar();
while(c < '0' || c > '9')
{
c = getchar();
}
while(c >= '0' && c <= '9')
{
x = x * 10 + (c - '0');
c = getchar();
}
return x;
}
int min (int x, int y)
{
return x < y ? x : y;
}
int max (int x, int y)
{
return x > y ? x : y;
}
int main ()
{
int T = 0, n = 0;
T = read();
while(T--)
{
n = read();
for(int i = 1;i <= n; i++)
{
a[i][0] = read();
a[i][1] = inf, a[i][2] = -inf;
}
if(!(n & 1))
{
printf("NO\n");
continue;
}
int head = 1, tail = 1;
f[head] = 1;
bool flag = false;
for(int i = 2;i <= n; i += 2)
{
while(head <= tail)
{
int x = f[head];
int L = a[x][1], R = a[x][2];
int lSonl = a[x][2];
int lSonr = min(a[x][1], a[x][0]);
int rSonl = max(a[x][2], a[x][0]);
int rSonr = a[x][1];
if(a[i][0] < lSonl || a[i][0] > lSonr || a[i + 1][0] < rSonl || a[i + 1][0] > rSonr)
{
head++;
}
else
{
break;
}
}
if(head > tail)
{
printf("NO\n");
flag = true;
break;
}
int x = f[head++];
a[i][1] = min(a[x][1], a[x][0]);
a[i][2] = a[x][2];
a[i + 1][1] = a[x][1];
a[i + 1][2] = max(a[x][0], a[x][2]);
f[++tail] = i;
f[++tail] = i + 1;
}
if(!flag)
{
printf("YES\n");
}
}
return 0;
}
|