HDU 5444 Elven Postman 二叉排序树

HDU 5444

题意：给你一棵树的先序遍历，中序遍历默认是1...n，然后q个查询，问根节点到该点的路径（题意挺难懂，还是我太傻逼）

思路:这他妈又是个大水题，可是我还是太傻逼。1000个点的树，居然用标准二叉树结构来存点，，，卧槽想些什么东西。可以用一维数组，left,right直接指向位置就行了，我这里用的是链表。对于读入的序列，生成一个二叉排序树，同时记录一下路径就可以了，后面居然忘了清空path又wa了一发。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #include<vector>
 #include<set>
 #include<string>
 #define inf 0x3f3f3f3f
 #define LL long long
 #define MAXN 2005
 #define IN freopen("in.txt","r",stdin);
 using namespace std;

 struct Node{
     int x;
     Node *left, *right;
     Node(int x = ) :x(x){
         left = NULL;
         right = NULL;
     };
 };

 vector<char> path[MAXN];

 Node *root;
 void bst_insert(int x){
     //Node *t = root;
     //return;
     Node *t = root;
     //Node *s = &Node(x);
     while (t != NULL){
         if (x > t->x){
             path[x].push_back('W');
             if (t->right == NULL){
                 t->right = new Node(x);
                 break;
             }
             else{
                 t = t->right;
             }
         }
         else{
             path[x].push_back('E');
             if (t->left == NULL){
                 t->left = new Node(x);
                 break;
             }
             else{
                 t = t->left;
             }
         }
     }
 }

 int main(){
     //IN;
     //freopen("out.txt", "w", stdout);
     int t, n;
     scanf("%d", &t);
     while (t--){
         scanf("%d", &n);
         int x;
         scanf("%d", &x);
         root = new Node(x);
         for (int i = ; i <= n; i++){
             path[i].clear();
         }
         for (int i = ; i <= n; i++) {
             scanf("%d", &x);
             bst_insert(x);
         }
         //    find(1, n, 1);
         int q;
         scanf("%d", &q);
         //bfs();
         while (q--){
             scanf("%d", &x);
             for (int i = ; i < path[x].size(); i++){
                 printf("%c", path[x][i]);
             }
             printf("\n");
         }
     }
     return ;
 }