题目：
输入两棵二叉树A和B，判断B是不是A的子结构。（ps：我们约定空树不是任意一个树的子结构）
过去提交通过的代码： 
class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }
}
public class Solution {
    public boolean HasSubtree(TreeNode root1,TreeNode root2) {         boolean result=false;         if(root1!=null&&root2!=null) {             if(root1.val==root2.val) {                 result=DoesRoot1HaveRoot2(root1,root2);             }             if(!result) {                 result=HasSubtree(root1.left,root2);             }             if(!result) {                 result=HasSubtree(root1.right,root2);             }         }         return result;
    }     boolean DoesRoot1HaveRoot2(TreeNode pRoot1,TreeNode pRoot2) {         if(pRoot2==null) {             return true;         }         if(pRoot1==null) {             return false;         }         if(pRoot1.val!=pRoot2.val) {             return false;         }         return DoesRoot1HaveRoot2(pRoot1.left,pRoot2.left)&&DoesRoot1HaveRoot2(pRoot1.right,pRoot2.right);     }
}
再写一次： 
/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {     public boolean HasSubtree(TreeNode root1,TreeNode root2) {         boolean res=false;         if (root1!=null&&root2!=null) {             if (root1.val==root2.val) {                 res=DoesRootHaveRoot2(root1,root2);             }             if (!res) {                 res=DoesRootHaveRoot2(root1.left,root2);             }             if (!res) {                 res=DoesRootHaveRoot2(root1.right,root2);             }         }         return res;     }     public boolean DoesRootHaveRoot2(TreeNode pRoot1,TreeNode pRoot2) {         if (pRoot2==null) {             return true;         }         if (pRoot1==null) {             return false;         }         if (pRoot1.val!=pRoot2.val) {             return false;         }         return DoesRootHaveRoot2(pRoot1.left,pRoot2.left)&&DoesRootHaveRoot2(pRoot1.right,pRoot2.right);     }
}