树的子结构

 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 DFS(TreeNode* pRoot1, TreeNode* pRoot2)
13     {
14         if ( pRoot1->val != pRoot2->val)
15             return 0;
16         bool b1,b2;
17         if (pRoot1->left != NULL && pRoot2->left != NULL)
18         {
19             b1 = DFS(pRoot1->left,pRoot2->left);
20         }
21         else if (pRoot2->left == NULL)
22         {
23             b1 = 1;
24         }
25         else
26         {
27             b1 = 0;
28         }
29 
30         if (pRoot1->right != NULL && pRoot2->right != NULL)
31         {
32             b2 = DFS(pRoot1->right,pRoot2->right);
33         }
34         else if (pRoot2->right == NULL)
35         {
36             b2 = 1;
37         }
38         else
39         {
40             b2 = 0;
41         }
42 
43         return b1 && b2;
44     }
45     bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
46     {
47         if (pRoot1 == NULL || pRoot2 == NULL)
48             return 0;
49         std::queue<TreeNode*> qq;
50         qq.push(pRoot1);
51         bool is = 0;
52         while(!qq.empty())
53         {
54             TreeNode* tem = qq.front();
55             qq.pop();
56             if(is = DFS(tem,pRoot2))
57                 return 1;
58             if (tem->left != NULL)
59                 qq.push(tem->left);
60             if (tem->right != NULL)
61                 qq.push(tem->right);
62         }
63         return 0;
64     }
65 };