The number of sub-structure to prove safety offer--

Description Title
input two binary A, B, B is judged not substructure A's. (Ps: we agreed empty tree is not a tree of any sub-structure).
Idea, two steps,
1 to find the root node of the same node B in A, referred to as A1 tree
2. Analyzing A1 contains B,

public class Main13 {
    public class TreeNode {
        int val = 0;
        TreeNode left = null;
        TreeNode right = null;
        public TreeNode(int val) {
            this.val = val;
        }
    }

    public boolean subtree(TreeNode root1,TreeNode root2){
        if (root1==null&&root2==null)//如果root2为空，说明B已经匹配完了，返回真
            return true;
        else if (root1!=null&&root2==null)//如果root2为空，说明B已经匹配完了，返回真
            return  true;
        else if (root1==null&&root2!=null)//如果root1先为空，说明B中的节点，A却没有，返回假
            return false;
        else {//如果都不为空
            if (root1.val!=root2.val){//判断值是否相等
                return false;
            }
            return subtree(root1.left,root2.left)&&subtree(root1.right,root2.right);//分别判断左右两边是否也是一样的
        }
    }
    
    public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        boolean flag=false;
       if (root1!=null&&root2!=null){
           if (root1.val==root2.val){//如果A中当前的节点和B的根节点相同,
               flag= subtree(root1,root2);//则判断A的子树是否和B相同
           }
           if (flag==false){//如果没找到，就看A的左节点是否和B的根节点相同
               flag=HasSubtree(root1.left,root2);
           }
           if(flag==false){//如果还没找到，就看A的右节点是否和B的根节点相同
               flag=HasSubtree(root1.right,root2);
           }
       }
       return flag;
    }
}