C++ 代码

class Solution {
public:

    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
       string str;
       dfs_s(root, str);
       return str;
    }
    void dfs_s(TreeNode* root, string &str){
       if(root == nullptr) {
            str += "# ";
            return;
        }

        str += to_string(root->val) + ' ';
        dfs_s(root->left, str);
        dfs_s(root->right, str);
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        TreeNode* root = nullptr;
        int i = 0;
        dfs1(root, data, i);
        return root;
    }
    string temp;//用这个进行转换
    void dfs1(TreeNode*& root, string& str, int& i){
        if(str[i] == '#'){
            i += 2;
            return;
        }
        if(i >= str.size()) return;
        int k = i;
        while(k < str.size() && str[k] != ' ')  ++k;
        for(int j = i; j < k; ++j)  temp += str[j];

        //直接用库函数的atoi就不用自己实现了判断正负什么的
        root = new TreeNode(atoi(temp.c_str()));

        temp.clear();
        i = k + 1;
        dfs1(root->left, str, i);
        dfs1(root->right, str, i);
    }
};