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:
Input: The root of a Binary Search Tree like this:
5
/ \
2 13
Output: The root of a Greater Tree like this:
18
/ \
20 13
Solution：
思路：
将中序遍历左-根-右的顺序逆过来，变成右-根-左的顺序，这样就可以反向计算累加和sum，同时更新结点值.
Time Complexity: O(N) Space Complexity: O(N)
Solution Code：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
int sum = 0;
public TreeNode convertBST(TreeNode root) {
convert(root);
return root;
}
public void convert(TreeNode cur) {
if (cur == null) return;
convert(cur.right);
cur.val += sum;
sum = cur.val;
convert(cur.left);
}
}
