Convert BST to Greater Tree
題意：
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.
Example:
The root of a Binary Search Tree like this:
5
/ \
2 13
Output:
The root of a Greater Tree like this:
18
/ \
20 13
解法：
由於只有比該node的值大的node，才會影響到該node的值，因此我們可以利用Reverse Inorder Traversal來遍歷一整棵樹，並維護一個Sum值，該Sum值即是紀錄著當下node之前所有比該node大的所有node的值之和。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
int sum = 0;
public TreeNode convertBST(TreeNode root) {
convert(root);
return root;
}
public void convert(TreeNode root) {
if (root == null) {
return;
}
convert(root.right);
root.val = sum + root.val;
sum = root.val;
convert(root.left);
}
}
Complexity:
Time: O(N)，N為node數量
Space: O(1)，因為我們只需一個sum值並且我們是直接修改原本的BST。