I like to read this solution as a small machine: keep the useful information, throw away the noise. For 297. Serialize and Deserialize Binary Tree, the solution in this repository is mainly a two pointers solution.
Guide
What?
The code is easier to read if we treat it as a controlled search through possible states. Instead of trying to be clever immediately, read the code as a sequence of questions:
- What state are we keeping?
- How do we move from one state to the next?
- When do we know the answer is already determined?
For this file, the main tools are: two pointers, sliding window.
Guide
When?
Reach for this shape when a direct simulation would work logically but waste time revisiting the same information. The accepted code reduces that pressure by storing exactly the information that remains useful later.
The important function names to track are join, string_split, serialize, deserialize, nodes.
Guide
Why?
The win comes from making each line carry responsibility: store the useful state, discard the rest, keep moving.
- The queue gives the solution a level-by-level or frontier-style traversal.
- Substring checks are convenient but not free, so they are part of the real complexity story.
- The final return is not magic; it is the invariant after the loops or recursion have finished doing their accounting.
Guide
How?
Walk through the solution in this order:
- Initialize the memory or helper structure.
- Process candidates in the order the invariant expects.
- Update the answer only when the current state is valid.
- Return the value that represents the fully processed input.
The most important competitive-programming habit here is to trust the invariant. Once the invariant is right, the loops become much less scary.
Guide
Complexity
- Time: O(n) to O(n log n), depending on the dominant loop or data structure operation
- Space: O(n) in the usual case for auxiliary containers or recursion
Guide
C++ Solution
Your submission
The accepted solution
01//Runtime: 452 ms, faster than 5.05% of C++ online submissions for Serialize and Deserialize Binary Tree.
02//Memory Usage: 38.7 MB, less than 41.53% of C++ online submissions for Serialize and Deserialize Binary Tree.
03/**
04 * Definition for a binary tree node.
05 * struct TreeNode {
06 * int val;
07 * TreeNode *left;
08 * TreeNode *right;
09 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
10 * };
11 */
12
13class Codec {
14public:
15 template <typename Iter>
16 std::string join(Iter begin, Iter end, std::string const& separator)
17 {
18 std::ostringstream result;
19 result.precision(2); //for floating point
20 if (begin != end)
21 result << *begin++;
22 while (begin != end)
23 //std::fixed : for floating point
24 result << std::fixed << separator << *begin++;
25 return result.str();
26 }
27
28 std::vector<std::string> string_split(std::string str, std::string delimiter){
29 size_t pos = 0;
30 std::string token;
31 std::vector<std::string> result;
32
33 //Style 2, can deal with all tokens inside the while loop
34 while(true){
35 pos = str.find(delimiter);
36 //works even if pos is string::npos
37 token = str.substr(0, pos);
38 result.push_back(token);
39 if(pos == string::npos) break;
40 //pos+1 equals to 0, so the line below can't handle this situation
41 str.erase(0, pos+delimiter.length());
42 }
43 return result;
44 }
45
46 // Encodes a tree to a single string.
47 string serialize(TreeNode* root) {
48 vector<string> tokens;
49
50 TreeNode* cur;
51 queue<TreeNode*> q;
52 q.push(root);
53
54 while(!q.empty()){
55 cur = q.front(); q.pop();
56
57 if(cur != nullptr){
58 tokens.push_back(to_string(cur->val));
59 //we also visit children who are nullptr
60 q.push(cur->left);
61 q.push(cur->right);
62 }else{
63 tokens.push_back("null");
64 }
65 }
66
67 return join(tokens.begin(), tokens.end(), ",");
68 }
69
70 // Decodes your encoded data to tree.
71 TreeNode* deserialize(string data) {
72 vector<string> tokens = string_split(data, ",");
73 int n = tokens.size();
74 vector<TreeNode*> nodes(n);
75
76 for(int i = 0; i < n; ++i){
77 if(tokens[i] == "null"){
78 nodes[i] = nullptr;
79 }else{
80 nodes[i] = new TreeNode(stoi(tokens[i]));
81 }
82 }
83
84 //use bfs to reconstruct tree
85 TreeNode* cur;
86 int nodeCursor = 0;
87 queue<TreeNode*> q;
88 q.push(nodes[nodeCursor]);
89
90 while(!q.empty()){
91 int levelSize = q.size();
92
93 //visit level by level
94 while(levelSize-- > 0){
95 cur = q.front(); q.pop();
96
97 if(cur != nullptr){
98 cur->left = nodes[++nodeCursor];
99 cur->right = nodes[++nodeCursor];
100
101 q.push(cur->left);
102 q.push(cur->right);
103 }
104 }
105 }
106
107 return nodes[0];
108 }
109};
110
111// Your Codec object will be instantiated and called as such:
112// Codec codec;
113// codec.deserialize(codec.serialize(root));
114
115//recursion, preorder
116//https://leetcode.com/problems/serialize-and-deserialize-binary-tree/discuss/74253/Easy-to-understand-Java-Solution
117//Runtime: 624 ms, faster than 5.05% of C++ online submissions for Serialize and Deserialize Binary Tree.
118//Memory Usage: 38.2 MB, less than 42.14% of C++ online submissions for Serialize and Deserialize Binary Tree.
119class Codec {
120public:
121 void serial(TreeNode* node, string& str){
122 if(node == nullptr){
123 str += "null";
124 }else{
125 str += to_string(node->val) + ",";
126 serial(node->left, str); str += ",";
127 serial(node->right, str);
128 }
129 }
130
131 // Encodes a tree to a single string.
132 string serialize(TreeNode* root) {
133 string str = "";
134 serial(root, str);
135 // cout << "str: " << str << endl;
136 return str;
137 }
138
139 std::vector<std::string> string_split(std::string str, std::string delimiter){
140 size_t pos = 0;
141 std::string token;
142 std::vector<std::string> result;
143
144 //Style 2, can deal with all tokens inside the while loop
145 while(true){
146 pos = str.find(delimiter);
147 //works even if pos is string::npos
148 token = str.substr(0, pos);
149 result.push_back(token);
150 if(pos == string::npos) break;
151 //pos+1 equals to 0, so the line below can't handle this situation
152 str.erase(0, pos+delimiter.length());
153 }
154 return result;
155 }
156
157 TreeNode* deserial(queue<string>& q){
158 string token = q.front(); q.pop();
159 if(token == "null") return nullptr;
160 TreeNode* root = new TreeNode(stoi(token));
161 root->left = deserial(q);
162 root->right = deserial(q);
163 return root;
164 };
165
166 // Decodes your encoded data to tree.
167 TreeNode* deserialize(string data) {
168 vector<string> tokens = string_split(data, ",");
169 queue<string, deque<string>> q(deque<string>(tokens.begin(), tokens.end()));
170 return deserial(q);
171 }
172};
Cost