## Sword Finger Offer 37-Serialized Binary Tree C++

Second question of debt repayment

## Solution BFS

I won't say much about BFS. Personally, the difficulty of this question is the processing between string and int.
First put a blog post C, C++ string summary to see the character summary.
This is the second time I have encountered ostringstream / istringstream. It is a very useful two input stream. The str() function of ostringstream can separate the one with a space. The serial numbers are converted to comma-separated strings
** The usage of isringstream in this question is still the same as that encountered in leetcode-58 : each string separated by commas in the large string is used as a new string Processing, the
format is probably like this

`````` 		istringstream in(data);
string tmp;
while(in >> tmp) {

//do something
}
``````

I also saw the stoi function: you can convert a string to a number. If you exceed the upper limit of int, you will get an error. More detailed knowledge
about c++ will be broken one by one during this winter vacation. After all, the interview is not just about writing a function. Fighting is much more than algorithms

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Codec {

public:

// Encodes a tree to a single string.
string serialize(TreeNode* root) {

ostringstream out;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()) {

TreeNode* tmp = q.front();
q.pop();
if(tmp != NULL) {

out<<tmp -> val<<" ";
q.push(tmp -> left);
q.push(tmp -> right);
}
else out<<"null ";
}
return out.str();
}

// Decodes your encoded data to tree.
TreeNode* deserialize(string data) {

istringstream in(data);
string tmp;
vector<TreeNode*> a;
while(in >> tmp) {

if(tmp =="null") {

a.push_back(NULL);
}
//stoi 把字符串转为int
else a.push_back(new TreeNode(stoi(tmp)));
}
int index = 1;
for(int i = 0; index < a.size(); i ++) {

if(a[i] == NULL) continue;
if(index < a.size()) a[i] -> left = a[index ++];
if(index < a.size()) a[i] -> right= a[index ++];
}
return a[0];
}
};

// Your Codec object will be instantiated and called as such:
// Codec codec;
// codec.deserialize(codec.serialize(root));
``````

Time complexity I don’t know how much ostringstream and isringstream are... Leave a hole in
space complexity

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Origin blog.csdn.net/qq_42883222/article/details/112984375
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