面试题37. 序列化二叉树

Posted 2020-11-29 ocpc

题目：

 

 

解答：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Codec {
11 public:
12 
13     // Encodes a tree to a single string.
14     string serialize(TreeNode* root) 
15     {
16         string res;
17         queue<TreeNode*> q;
18         q.push(root);
19 
20         while (!q.empty()) 
21         {
22             auto node = q.front();
23             if (node) 
24             {
25                 res += to_string(node->val) + ‘ ‘;
26                 q.push(node->left);
27                 q.push(node->right);
28             } 
29             else 
30             {
31                 res += "null ";
32             }
33             q.pop();
34         }
35 
36         // cout << res;
37         return res;
38     }
39     
40     // Decodes your encoded data to tree.
41     TreeNode* deserialize(string data) 
42     {
43         vector<TreeNode*> list;
44         for (int l = 0, r = 0; l < data.size(); ) 
45         {
46             r = l;
47             while (data[r] != ‘ ‘) 
48             {
49                 r ++ ;
50             }
51             if (data[l] == ‘n‘) 
52             {
53                 l = r + 1;
54                 list.push_back(NULL);
55                 continue;
56             }
57             int val = 0, sign = 1;
58             if (l < r && data[l] == ‘-‘) 
59             {
60                 sign = -1, l ++ ;
61             }
62             for (int i = l; i < r; i ++ ) 
63             {
64                 val = val * 10 + data[i] - ‘0‘;
65             }
66             val *= sign;
67             l = r + 1;
68             auto newNode = new TreeNode(val);
69             list.push_back(newNode);
70         }
71         
72         auto newList = list[0];
73         auto curNode = newList;
74         
75         // j表示当前的根节点，遇到null就递进
76         // i表示当前的孩子节点，相邻两个构成一个父节点的孩子节点
77         for (int i = 1, j = 1; j < list.size() && i < list.size(); j ++ ) 
78         {
79             if (!curNode) 
80             {
81                 curNode = list[j];
82                 continue;
83             }
84             curNode->left = list[i];
85             curNode->right = list[i + 1];
86             curNode = list[j];
87             i += 2;
88         }
89 
90         return newList;
91     }
92 };