PAT L3-016：二叉搜索树的结构（暴力）

https://www.patest.cn/contests/gplt/L3-016

题意：中文。

思路：暴力构造，暴力查询就好了。只不过操作很多，很麻烦。比赛的时候再给我10分钟就打完了，手速太慢好可惜。

 #include <bits/stdc++.h>
 using namespace std;
 #define N 100010
 typedef long long LL;
 struct Node {
     int num;
     Node *l, *r, *fa;
     Node (int n) {
         l = r = fa = NULL;
         num = n;
     }
 } ;
 Node *root;
 int flag, d;
 char s[][];

 void Insert(Node *now, Node *f, int x, int kind) {
     if(now == NULL) {
         now = new Node(x);
         now->fa = f;
         if(kind) f->r = now;
         else f->l = now;
         return ;
     }
     if(x > now->num) Insert(now->r, now, x, );
     else Insert(now->l, now, x, );
 }

 Node *Find(int x) {
     Node *now = root; d = ;
     while(now != NULL && now->num != x) {
         if(x > now->num) now = now->r;
         else now = now->l;
         d++;
     }
     return now;
 }

 void Solve1(int x) {
     if(root && root->num == x) flag = ;
 }

 void Solve2(int x, int y) {
     Node *now1 = Find(x), *now2 = Find(y);
     if(now1 && now2 && now1->fa == now2->fa) flag = ;
 }

 void Solve3(int x, int y) {
     Node *now = Find(y);
     if(now && now->fa && now->fa->num == x) flag = ;
 }

 void Solve4(int x, int y) {
     Node *now = Find(y);
     if(now && now->l && now->l->num == x) flag = ;
 }

 void Solve5(int x, int y) {
     Node *now = Find(y);
     if(now && now->r && now->r->num == x) flag = ;
 }

 void Solve6(int x, int y) {
     Node *now1 = Find(x);
     int d1 = d;
     Node *now2 = Find(y);
     int d2 = d;
     if(now1 && now2 && d1 == d2) flag = ;
 }

 int main() {
     int n, a, b; scanf("%d", &n);
     if(n) scanf("%d", &a), root = new Node(a);
     for(int i = ; i < n; i++) scanf("%d", &a), Insert(root, NULL, a, );
     int q; scanf("%d", &q);
     while(q--) {
         flag = ;
         scanf("%d", &a);
         scanf("%s", s[]);
         if(s[][] == 'a') {
             scanf("%d", &b);
             scanf("%s %s", s[], s[]);
             if(s[][] == 's') {
                 Solve2(a, b);
             } else {
                 Solve6(a, b);
                 scanf("%s %s %s", s[], s[], s[]);
             }
         } else {
             scanf("%s %s", s[], s[]);
             if(s[][] == 'o') {
                 Solve1(a);
             } else if(s[][] == 'a') {
                 scanf("%s %d", s[], &b), Solve3(a, b);
             } else if(s[][] == 'e') {
                 scanf("%s %s %d", s[], s[], &b), Solve4(a, b);
             } else {
                 scanf("%s %s %d", s[], s[], &b), Solve5(a, b);
             }
         }
         if(flag) puts("Yes");
         else puts("No");
     }
     return ;
 }