Record the reason for "memory limit exceeded" once

question:

        The source of the problem is this question from LeetCode:LCR 048. Serialization and deserialization of binary trees - LeetCode

        I thought it was okay, just do preorder traversal. The time and space complexity are both O(n), so I should be able to get rid of the problem K.

Question code:

        

/**
 * 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:
    list<string> str2lst(string str)    
    {
        list<string> ret;
        string cur = "";
        for(int i =0;i<str.size();++i){
           if(str[i] ==' '){
               if(cur.size()>0){
                   ret.push_back(cur);
                   cur ="";
               }
               continue;
           }
           cur = cur + str[i];
        }
        return ret;
    }
    TreeNode *  dfs2(list<string> & arr) //反序列化
    {
        if(arr.front()=="*"){
            arr.erase(arr.begin());
            return nullptr;
        }
        TreeNode * cur = new TreeNode(stoi(arr.front()));
        arr.erase(arr.begin());
        cur->left = dfs2(arr);
        cur->right = dfs2(arr);
        return cur;
    }
    void dfs(TreeNode * root,string & s){
        if(root == nullptr){
            s.push_back('*');
            s =s + " ";
            return;
        }
        s=s+(to_string(root->val)+" ");
        dfs(root->left,s);
        dfs(root->right,s);
    }
    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        string ret = "";
        dfs(root,ret);
        return ret;
    
    }
    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        if(data.size()==0){
            return nullptr;
        }
        list<string> arr = str2lst(data);
        return dfs2(arr);
    }
};

result:

       

Stuck example:

(However, the amount of data does not exceed the scope of the question)

And I compared it with the official solution, and the ideas are almost exactly the same.

But my code times out. . . .

analyze:

        So how could I let go of such a weird event? Then I modified my code bit by bit to run the test to see where the problem was.

        Finally found the code of the problem:

s=s+(to_string(root->val)+" ");

change it to

s+=(to_string(root->val)+" ");

The code will pass. . . . . .

Because if you use the former, a new temporary string object will be created during the execution process, which will double the memory usage and consume a lot of time (especially when s here is actually very large, the additional time and space costs will be higher).

so,

The "higher" here may actually make a huge difference.

In the usual development process, you should develop the habit of using more efficient syntax.