剑指Offer——序列化二叉树

 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 */
11 class Solution {
12 public:
13     vector<int> buf;
14     void dfs1(TreeNode *root) {
15         if(!root) buf.push_back(0xFFFFFFFF);
16         else {
17             buf.push_back(root->val);
18             dfs1(root->left);
19             dfs1(root->right);
20         }
21     }
22     TreeNode* dfs2(int* &p) {
23         if(*p == 0xFFFFFFFF) {
24             p++;
25             return NULL;
26         }
27         TreeNode* res = new TreeNode(*p);
28         p++;
29         res->left = dfs2(p);
30         res->right = dfs2(p);
31         return res;
32     }
33     char* Serialize(TreeNode *root) {
34         buf.clear();
35         dfs1(root);
36         int bufSize = buf.size();
37         int *res = new int[bufSize];
38         for(int i = 0; i < bufSize; i++) res[i] = buf[i];
39         return (char*)res;
40     }
41     TreeNode* Deserialize(char *str) {
42         int *p = (int*)str;
43         return dfs2(p);
44     }
45 };