树的同构 java

import java.util.Scanner;
//创造节点类型
class Node 
{
    String key;
    int left;
    int right;
}
class test
{
    public static void main(String[] args)
    {
        Scanner sc=new Scanner(System.in);

        Node[] arr1=new Node[8];
        Node[] arr2=new Node[8];

        int head1=biuldTree(arr1, sc);
        int head2=biuldTree(arr2, sc);
        

        boolean result=isTonggou(arr1,head1,arr2,head2);
        if (result)
        {
            System.out.print("Yes");
        }
        else
        {
            System.out.print("No");
        }
    }
    public static int biuldTree(Node[] arr,Scanner sc)
    {
        int N=sc.nextInt();
        String templeft;
        String tempright;
        int[] flag=new int[N];
        int root=0;
        
        for (int i=0;i<N ;i++ )
        {
            Node node=new Node();
            node.key=sc.next();
            templeft=sc.next();
            tempright=sc.next();
            arr[i]=node;
            

            if (templeft.equals("-"))
            {
                node.left=-1;
            }
            else
            {
                node.left=Integer.parseInt(templeft);
                flag[node.left]=1;
            }
            if (tempright.equals("-"))
            {
                node.right=-1;
            }
            else
            {
                node.right=Integer.parseInt(tempright);
                flag[node.right]=1;
            }
            for (int j=0;j<N ;j++ )
            {
                if (flag[j]==0)
                {
                    root=j;
                }
            }
            


        }
        return root;
    }
    public static boolean isTonggou(Node[] arr1,int head1,Node[] arr2,int head2)
    {
        //递归出口
        if ((head1==-1) &&( head2==-1))
        {return true;}
        if ((head1==-1) && (head2 !=-1) || (head1!=-1) && (head2==-1))
        {return false;}
        if (!(arr1[head1].key.equals(arr2[head2].key)))
        {return false;}
        //树1和树2的左孩子都为空，判断右孩子为根节点的树是不是同构
        if ((arr1[head1].left==-1) && (arr2[head2].left==-1))
        {
            return isTonggou(arr1,arr1[head1].right, arr2, arr2[head2].right);
        }
        //如果树1的左孩子不空，树2的左孩子也不空，并且两个左孩子的key相等，那么判断以左孩子为根的树是否同构，以右孩子为根的树是否同构
        if ((arr1[head1].left != -1 )&&( arr2[head2].left != -1 )&& (arr1[arr1[head1].left].key.equals((arr2[arr2[head2].left].key)) ))
        {
            return (isTonggou(arr1, arr1[head1].left,arr2, arr2[head2].left)&&(isTonggou(arr1,arr1[head1].right,arr2,arr2[head2].right)));

        }
        //如果上述都不满足，那么判断以树1的左孩子和树2的右孩子为根的树是否同构，判断以树1的右孩子和树2的左孩子为根的树是否同构
        else
        {
            return (isTonggou(arr1, arr1[head1].left,arr2, arr2[head2].right)&&(isTonggou(arr1,arr1[head1].right,arr2,arr2[head2].left)));

        }
    }
}

实验结果

采用结构数组来储存节点，将孩子为空标记为-1.

Yes和No前面的树分别标记两颗树的根所在的位置，是我调试时加上以便判断哪个地方出问题的。