class Solution {

    // 先將右子樹轉換為累加樹；並記錄右子樹的累加和rightSum;
    // 然後處理根節點，根節點的值 = 根節點值 + rightSum;
    // 然後轉化左子樹
    public TreeNode convertBST(TreeNode root) {
        convert(root, 0);
        return root;
    }
    
    public int convert(TreeNode root, int rightSum){
        if(null == root){
            return rightSum;
        }
        rightSum = convert(root.right, rightSum);
        root.val = root.val + rightSum;
        rightSum = root.val;
        rightSum = convert(root.left, rightSum);
        return rightSum;
    }
}
 

class Solution {
    
    // !!! 不太很優雅，但卻讓程式碼更容易理解!!!
    private int rightSum = 0;

    // 先將右子樹轉換為累加樹；並記錄右子樹的累加和rightSum;
    // 然後處理根節點，根節點的值 = 根節點值 + rightSum;
    // 然後轉化左子樹
    public TreeNode convertBST(TreeNode root) {
        if(null == root){
            return root;
        }
        convertBST(root.right);
        root.val = root.val + rightSum;
        rightSum = root.val;
        convertBST(root.left);
        return root;
    }
}