树的同构--静态链表建树

不熟悉静态链表利用左右子节点下标建树，以及如何寻找根节点，使用静态链表建树传入根节点只要传入根节点下标。

#include<iostream>
#include<algorithm>
#include<stack>
#include<cstring>
using namespace std;
struct node
{
	char data;
	int left;
	int right;
}T1[10],T2[10];
bool check[10];
int buildTree(node T[])
{
	int root=-1, i;
	int n; cin >> n;   
	for(int i=0; i<n; i++)check[i]=0;
	char a, b;
	for (i = 0; i < n; i++)
	{
		cin >> T[i].data >> a>> b;
		if (a != '-')
		{
			T[i].left = a - '0';
			check[T[i].left] = 1;
		}	
		else
			T[i].left = -1;
		if (b != '-')
		{
			T[i].right = b - '0';
			check[T[i].right] = 1;
		}
		else
			T[i].right = -1;
	}
	for (i = 0; i < n; i++)
	{
		if (check[i] != 1)
		{
			root = i;
			break;
		}
	}
	return root;
}
bool judge(int R1, int R2) {
	if (R1 == -1 && R2 == -1) 
		return true;
	if (R1 == -1 && R2 != -1 || R1 != -1 && R2 == -1)   
		return false;
	if (T1[R1].data != T2[R2].data) 
		return false;
	return (judge(T1[R1].left, T2[R2].left) && 
		judge(T1[R1].right, T2[R2].right))||
		(judge(T1[R1].right, T2[R2].left) && 
		judge(T1[R1].left, T2[R2].right));
}
int main()
{
	int r1, r2;
	r1 = buildTree(T1);
	r2 = buildTree(T2);
	if (judge(r1, r2))
		cout << "Yes";
	else
		cout << "No";
	return 0;
}