题目：

Given a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.

思路：

每个结点值加上所有比它大的结点值总和当作新的结点值。

初始化sum值 = 0

因为inorder traversal （left->root->right）结果是non decreasing 递增数列，那么逆过来right->root->left,就可以得到递减数列, 每遍历到一个结点更新sum值，并将更新后的sum值赋给当前结点，从而保证，在当前结点，所有比他大的结点都访问过了，并且sum值就是所有比他大的结点值的总合。

代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int sum = 0;
    
    public TreeNode convertBST(TreeNode root) {
        dfs(root);
        return root;
    }
    private void dfs(TreeNode root) {
        if(root == null) return;
        dfs(root.right);
        sum += root.val;
        root.val = sum;
        dfs(root.left);
    }
}