今天上班搞一整天，只解出一個BUG，結果下班以後腦袋比較靈光?
總之今天是順利解出來了

題號:129 標題：Sum Root to Leaf Numbers 難度：Medium

You are given the root of a binary tree containing digits from 0 to 9 only.
Each root-to-leaf path in the tree represents a number.
• For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123.
Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.
A leaf node is a node with no children.

Example 1:

Input: root = [1,2,3]
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.
Example 2:

Input: root = [4,9,0,5,1]
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

Constraints:
• The number of nodes in the tree is in the range [1, 1000].
• 0 <= Node.val <= 9
• The depth of the tree will not exceed 10.

我的程式碼

class Solution {
    public int sumNumbers(TreeNode root) {
        if(root == null){
            return 0;
        }else if(root.left==null && root.right==null){
            return root.val;
        }
        TreeNode current = root;
        int resultf = 0;
        return chk(root,resultf);          
    }
    
    public int chk(TreeNode now,int result) {
        result = result*10 +now.val;
        if(now.left==null && now.right == null){
            return result;
        }
        if(now.left==null){
            return chk(now.right,result);
        }else if(now.right==null){
            return chk(now.left,result);
        }
        return chk(now.left,result) + chk(now.right,result);
    }   
    
}

我的邏輯
把每一層的數字乘以十加上這一層的數字，如果有下一層，就往下一層找，重複這個邏輯。

DAY27心得
還剩3天，加油啊