題目：

TinyURL is a URL shortening service where you enter a URL such as https://leetcode.com/problems/design-tinyurl and it returns a short URL such as http://tinyurl.com/4e9iAk.
Design the encode and decode methods for the TinyURL service. There is no restriction on how your encode/decode algorithm should work. You just need to ensure that a URL can be encoded to a tiny URL and the tiny URL can be decoded to the original URL.

心得：

這題實務上的話我應該會用資料庫來存key，簡單易用不是嗎？
不過要小心的是不能使用流水號來當作tinyurl的參數，
這樣會有被猜出key的可能性，可能就隨機產生一個短的字串並檢查不能重複來當key

我的答案(轉base64)：

public class Codec {

    // Encodes a URL to a shortened URL
    public string encode(string longUrl) {
        return Convert.ToBase64String(Encoding.UTF8.GetBytes(longUrl));
    }

    // Decodes a shortened URL to its original URL.
    public string decode(string shortUrl) {
        return Encoding.UTF8.GetString(Convert.FromBase64String(shortUrl));
    }
}

我的答案(guid)：

public class Codec {
    
    private static Dictionary<string, string> dty = new Dictionary<string, string>();
    
    // Encodes a URL to a shortened URL
    public string encode(string longUrl) {
        string guid = Guid.NewGuid().ToString();
        dty.Add(guid, longUrl);
        return guid;
    }

    // Decodes a shortened URL to its original URL.
    public string decode(string shortUrl) {
        string url = string.Empty;
        return dty.TryGetValue(shortUrl, out url) ? url : null;
    }
}

問題：

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.

You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

心得：

這題就是把兩個二元樹合併，我就想著最簡單遞迴來解決，看了最佳解答才發現如果為一邊null的話就可以直接用另外一邊來return更有效率，學習了。
我的答案：

public class Solution {
    public TreeNode MergeTrees(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null)
        {
            return null;
        }

        int t1Val = (t1 == null) ? 0 : t1.val;
        int t2Val = (t2 == null) ? 0 : t2.val;

        return new TreeNode(t1Val + t2Val)
        {
            left = MergeTrees(t1?.left, t2?.left),
            right = MergeTrees(t1?.right, t2?.right)
        };
    }
}

最佳解答：

public class Solution {
    public TreeNode MergeTrees(TreeNode t1, TreeNode t2) {
        if (t1 == null)
        {
            return t2;
        }
        if (t2 == null)
        {
            return t1;
        }

        return new TreeNode(t1.val + t2.val)
        {
            left = MergeTrees(t1.left, t2.left),
            right = MergeTrees(t1.right, t2.right)
        };
    }
}