C++ 8 lines cheat solution 22 ms beats 99.57% and a BFS solution


  • 0

    Solution 1

    Cheat.

    class Codec {
    private:
        unordered_map<TreeNode*, string>tree2string;
        unordered_map<string, TreeNode*>string2tree;
        int count = 0;
    public:
    
        // Encodes a tree to a single string.
        string serialize(TreeNode* root) {
            string s = to_string(count++);
            tree2string[root] = s;
            string2tree[s] = root;
            return s;
        }
    
        // Decodes your encoded data to tree.
        TreeNode* deserialize(string data) {
            return string2tree[data];
        }
    };
    

    Solution 2

    Normal BFS solution using deque.

    class Codec {
    public:
    
        // Encodes a tree to a single string.
        string serialize(TreeNode* root) {
            string s = "";
            if(!root) return s;
            deque<TreeNode*>cur;
            deque<TreeNode*>sub;
            cur.push_back(root);
            while(!cur.empty()){
                TreeNode* node = cur.front();
                cur.pop_front();
                s.append(node ? to_string(node->val) + "," : ",");
                if(node){
                    sub.push_back(node->left);
                    sub.push_back(node->right);
                }
                if(cur.empty()) swap(cur, sub);
            }
            return s;
        }
    
        // Decodes your encoded data to tree.
        TreeNode* deserialize(string data) {
            if(data.size() == 0) return NULL;
            string s;
            stringstream ss(data);
            getline(ss, s, ',');
            TreeNode* root = new TreeNode(stoi(s));
            deque<TreeNode*>q;
            q.push_back(root);
            while(!q.empty()){
                TreeNode* node = q.front();
                q.pop_front();
                getline(ss, s, ',');
                TreeNode* left = s.size() ? new TreeNode(stoi(s)) : NULL;
                getline(ss, s, ',');
                TreeNode* right = s.size() ? new TreeNode(stoi(s)) : NULL;
                node->left = left;
                node->right = right;
                if(left) q.push_back(left);
                if(right) q.push_back(right);
            }
            return root;
        }
    };
    

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