7-2 是否同一棵二叉搜索树

给定一个插入序列就可以唯一确定一棵二叉搜索树。然而，一棵给定的二叉搜索树却可以由多种不同的插入序列得到。例如分别按照序列{2, 1, 3}和{2, 3, 1}插入初始为空的二叉搜索树，都得到一样的结果。于是对于输入的各种插入序列，你需要判断它们是否能生成一样的二叉搜索树。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

struct Node {
int val;
Node *left;
Node *right;
Node() { val = 0; left = right = NULL; }
};

struct BST {
Node *root;
int cnt, inorder[15];
void init() { root = new Node; cnt = 0; }
void insert(int x) {
Node *t = root, *fa = root;
while (t != NULL) {
fa = t;
//cout << (t->val) << endl;
if (x <= (t->val)) t = (t->left);
else t = (t->right);
}
// cout << "test" << endl;
t = new Node;
if (x <= fa->val) fa->left = t;
else fa->right = t;
t->val = x;
}
void solve(Node *t) {
if (t == NULL) return;
inorder[++cnt] = (t->val);
//cout << inorder[cnt] << endl;
solve(t->left);
solve(t->right);
}
}bst, temp;

int main() {
int n, m, t;
while (~scanf("%d", &n) && n) {
scanf("%d", &m);
bst.init();
for (int i = 1; i <= n; i++) {
scanf("%d", &t);
bst.insert(t);
}
bst.solve(bst.root);
// for (int i = 2; i <= n + 1; i++) {
// cout << bst.inorder[i] << "/";
// }
// cout << endl;
for (int i = 1; i <= m; i++) {
temp.init();
for (int j = 1; j <= n; j++) {
scanf("%d", &t);
temp.insert(t);
}
temp.solve(temp.root);
// for (int j = 2; j <= n + 1; j++) cout << temp.inorder[j] << "/";
// cout << endl;
bool ok = true;
for (int j = 2; j <= n + 1; j++) if (bst.inorder[j] != temp.inorder[j]) { ok = false; break; }
if (ok) printf("Yes\n");
else printf("No\n");
}
}
return 0;
}