Whether the same binary search tree (25 points)

Given an insertion sequence, a binary search tree can be uniquely determined . However , A given binary search tree can be obtained from a variety of different insertion sequences . For example, in sequence {2, 1, 3} and {2, 3, 1} Insert a binary search tree that is initially empty , All get the same result . So for the input of various insertion sequences , You need to determine if they can generate the same binary search tree .

Input format :

The input contains several sets of test data . The... Of each group of data 1 Line gives two positive integers N (≤10) and L, They are the number of inserted elements in each sequence and the number of sequences to be checked . The first 2 Line is given N Positive integers separated by spaces , As the initial insertion sequence . And then L That's ok , Each line gives N An inserted element , Belong to L A sequence to check .

Simplicity , We guarantee that every insertion sequence is 1 To N An arrangement of . When read N by 0 when , Mark end of input , Don't process this data .

Output format :

For each set of sequences that need to be checked , If the binary search tree generated by it is the same as that generated by the corresponding initial sequence , Output “Yes”, Otherwise output “No”.

sample input :

4 2
3 1 4 2
3 4 1 2
3 2 4 1
2 1
2 1
1 2
0

sample output :

Yes
No
No

#include<bits/stdc++.h>
using namespace std;
struct Node {
	int data;
	Node* left;
	Node* right;
};
typedef Node* Tree;
Tree insert(Tree BT, int x) {
	if (!BT) {
		BT = new (Node);
		BT->data = x;
		BT->left = BT->right = NULL;
	} else if (x < BT->data)
		BT->left = insert(BT->left, x);
	else if (x > BT->data)
		BT->right = insert(BT->right, x);
	return BT;
}
bool judge(Tree a, Tree b) {
	if (a == NULL && b == NULL)
		return true;
	else if ((a != NULL && b == NULL) || (a == NULL && b != NULL))
		return false;
	else if (a->data == b->data)
		return judge(a->left,b->left)&&judge(a->right,b->right);
	return false;
}
int main(){
	int n,l,k;
	while(cin>>n&&n){
		cin>>l;
		Tree BT=NULL;
		int x;
		for(int i=0;i<n;i++){
			cin>>x;
			BT=insert(BT,x);
		}
		for(int i=0;i<l;i++){
			Tree Temp=NULL;
			for(int j=0;j<n;j++){
				cin>>x;
				Temp=insert(Temp,x);
			}
			if(judge(Temp,BT))cout<<"Yes\n";
			else cout<<"No\n";
		}
	}
}