# Leetcode [No.606] Construct String from Binary Tree (EASY)

## 題目

https://leetcode.com/problems/construct-string-from-binary-tree/description/?envType=daily-question&envId=2023-12-12

## 思路
+ 這個題目在描述的時候有一點問題，我也是看了討論區才知道題目想要幹嘛...
+ 
+ 根據 `yogeshwarb`所描述的，我們只要在**有右子樹但無左子樹**的時候加一個()即可。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string tree2str(TreeNode* root) 
    {
        string ans = "";
        preorderTraverse(root, ans);
        return ans;
    }
    void preorderTraverse(TreeNode* node, string& ans)
    {
        if(node == nullptr) return;
        else
        {
            // cout << node->val;
            string str = to_string(node->val);
            ans+= str;
            if(node->left != nullptr)
            {
                // cout << "(";
                ans+= "(";
                preorderTraverse(node->left, ans);
                // cout << ")";
                ans+=")";
            }
            if(node->right != nullptr)
            {
                if(node->left == nullptr)
                {
                    ans+="()";
                }
                ans+="(";
                // cout << "(";
                preorderTraverse(node->right, ans);
                // cout << ")";
                ans+=")";
            }
        }

    }
};

### 解法分析
+ time complexity: O(lgn) // tree based

### 執行結果


---

## 改良:

+ 改成沒那麼白癡的版本，把left的()加在preorderTraverse的外面

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string tree2str(TreeNode* root) 
    {
        string ans = "";
        preorderTraverse(root, ans);
        return ans;
    }
    void preorderTraverse(TreeNode* node, string& ans)
    {
        if(node == nullptr) return;
        else
        {
            string str = to_string(node->val);
            ans+= str;
            if (node->left == nullptr && node->right == nullptr)
            {
                return;
            }
            else // there exist left or exist right
            {
                ans += "(";
                if(node->left)
                {
                    preorderTraverse(node->left, ans);
                }
                ans += ")";

                if(node->right)
                {
                    ans += "(";
                    preorderTraverse(node->right, ans);
                    ans += ")";
                }
            }
        }

    }
};

### 執行結果
會稍微慢一些..