Java如何实现树的同构?

public static class Node{
        String data;
        int left;
        int right;
         }

public class TongGou {

    private Scanner scanner;

    public TongGou(){
        scanner = new Scanner(System.in);
    }

    //树结构
    public static class Node{
        String data;
        int left;
        int right;

    }

    /**
     * 创建树
     * @param nodes
     * @return
     */
    public int createTree(Node[] nodes){
        int N = nodes.length;
        int root = -1;
        int[] check = new int[N];
        Arrays.fill(check,0);  //初始化为0
        for (int i=0;i<N;i++){
            //输入格式  data,left,right
            String next = scanner.next();
            String[] inputList = next!=null?next.split(","):null;
            if(inputList!=null&&inputList.length==3){
                nodes[i] = new Node();
                int  left = "-".equals(inputList[1])?-1:Integer.parseInt(inputList[1]);
                int  right = "-".equals(inputList[2])?-1:Integer.parseInt(inputList[2]);
                nodes[i].data = inputList[0];
                nodes[i].left = left;
                nodes[i].right = right;

                if(left>0) {
                    check[left] = 1;
                }
                if(right>0){
                    check[right] = 1;
                }

            }

        }

        for(int i=0;i<check.length;i++){
            if(check[i]==0&&nodes[i].data!=null){
                root = i;
                break;
            }
        }

        return root;
    }

    /**
     * 判断同构
     * @param r1
     * @param r2
     * @return
     */
    public boolean isomorphic(int r1,int r2,Node[] t1,Node[] t2){
        //须注意不要漏掉逻辑！
        
        //两个根节点均为null，必同构
        if ((r1 == -1) && (r2 == -1)) {
            return true;
        }
        //一个非空 另一个空，必不同构
        if(((r1==-1)&&(r2!=-1))||((r1!=-1)&&(r2==-1))){
            return false;
        }
        //两个节点非空 但值不同，必不同构
        if(!t1[r1].data.equals(t2[r2].data)){
            return false;
        }
        //两根节点的左孩子为空条件下，则须判断两根节点的右子树是否同构
        if(t1[r1].left==-1&&t2[r2].left==-1){
            return isomorphic(t1[r1].right,t2[r2].right,t1,t2);
        }
        //两根节点的左孩子不为空且左孩子的值也相同，须判断两根节点的左子树是否同构以及两根节点的右子树是否同构
        //如果左右子树均同构，则整棵树同构
        if((t1[r1].left!=-1&&t2[r2].left!=-1)&&(t1[t1[r1].left].data.equals(t2[t2[r2].left].data))){
            return isomorphic(t1[r1].left,t2[r2].left,t1,t2)&&isomorphic(t1[r1].right,t2[r2].right,t1,t2);
        }else{
            //分两种情况解释：
            //1、两根节点的左孩子不为空，但左孩子的值不同
            //例如：t1[r1.left].data!=t2[r2.left].data。但有t1[r1.left].data==t2[r2.right].data、t1[r1.right].data==t2[r2.left].data
            //即有可能r1的左子树与r2的右子树同构、r1的右子树与r2的左子树同构
            //故须判断r1的左子树是否与r2的右子树同构,以及r1的右子树是否与r2的左子树同构
            //2、两根节点的左孩子一个为空，一个不为空
            //例如：r1.left==-1、r2.left!=-1，如果r2.right==-1,显然r1的左子树与r2的右子树同构，此时则有可能r1的右子树与r2的左子树同构
            //故须判断r1的左子树是否与r2的右子树同构,以及r1的右子树是否与r2的左子树同构
            return isomorphic(t1[r1].left,t2[r2].right,t1,t2)&&isomorphic(t1[r1].right,t2[r2].left,t1,t2);
        }

    }

    public static void main(String[] args) {
        TongGou tongGou = new TongGou();
        Node[] nodes = new Node[4];
        Node[] nodes1 = new Node[4];
        int tree1 = tongGou.createTree(nodes);
        System.out.println();
        int tree2 = tongGou.createTree(nodes1);
        boolean isomorphic = tongGou.isomorphic(tree1, tree2, nodes, nodes1);
        System.out.println(isomorphic);

    }


}