18 树的子结构

输入两棵二叉树A，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）

C++:

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution {
11 public:
12     bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2){
13         bool result = false ;
14         if (pRoot1 != NULL && pRoot2 != NULL){
15             if (pRoot1->val == pRoot2->val)
16                 result = isSubtree(pRoot1 , pRoot2) ;
17             if (!result)
18                 result = HasSubtree(pRoot1->left , pRoot2) ;
19             if (!result)
20                 result = HasSubtree(pRoot1->right , pRoot2) ;
21         }
22         return result ;
23     }
24     
25     bool isSubtree(TreeNode* pRoot1, TreeNode* pRoot2){
26         if (pRoot2 == NULL)
27             return true ;
28         if (pRoot1 == NULL)
29             return false ;
30         if (pRoot1->val != pRoot2->val)
31             return false ;
32         return isSubtree(pRoot1->left , pRoot2->left) && isSubtree(pRoot1->right , pRoot2->right) ;
33     }
34 };