Given the root of 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 the sum of all keys greater than the original key in BST.

As a reminder, a binary search tree is a tree that satisfies these constraints:


	The left subtree of a node contains only nodes with keys less than the node's key.
	The right subtree of a node contains only nodes with keys greater than the node's key.
	Both the left and right subtrees must also be binary search trees.


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Example 1:

Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]


Example 2:

Input: root = [0,null,1]
Output: [1,null,1]


&nbsp;
Constraints:


	The number of nodes in the tree is in the range [0, 104].
	-104 &lt;= Node.val &lt;= 104
	All the values in the tree are unique.
	root is guaranteed to be a valid binary search tree.


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Note: This question is the same as 1038: https://leetcode.com/problems/binary-search-tree-to-greater-sum-tree/

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        ans = []
        def inorder(node):
            if not node:
                return node
            inorder(node.left)
            ans.append(node.val)
            inorder(node.right)
        def dfs(node):
            if not node:
                return None
            idx = ans.index(node.val)
            node.val = node.val + sum(ans[idx+1:])
            dfs(node.left)
            dfs(node.right)

        inorder(root)
        dfs(root)
        return root
        
class Solution {
    public TreeNode convertBST(TreeNode root) {
    if(root!=null) {


        List<Integer> nodesValues = new ArrayList<>();
        helperNodesVales(root, nodesValues);
        traverseAndAdd(root, nodesValues);

        return root;
    }
        return null;
    }

    private void helperNodesVales(TreeNode root, List<Integer> nodesValues) {
        if (root != null) {
            nodesValues.add(root.val);
        }
        if (root.right != null) {
            helperNodesVales(root.right, nodesValues);
        }
        if (root.left != null) {
            helperNodesVales(root.left, nodesValues);
        }
        if (root == null) {
            return;
        }
    }

    private void traverseAndAdd(TreeNode root, List<Integer> nodesValues) {
        if (root != null) {
            int rootVal = root.val;
            for (int i = 0; i < nodesValues.size(); i++) {
                if (nodesValues.get(i) > rootVal)
                    root.val += nodesValues.get(i);
            }
        }
        if (root.right != null) {
            traverseAndAdd(root.right, nodesValues);
        }
        if (root.left != null) {
            traverseAndAdd(root.left, nodesValues);
        }
        if (root == null) {
            return;
        }
    }
}
class Solution {
public:
    int val = 0;
    TreeNode* convertBST(TreeNode* root) {

        if(root)
        {
            convertBST(root->right); // traverse right sub-tree
            val += root->val; // add val
            root->val = val; // update val
            convertBST(root->left); // traverse left sub-tree
        }
        
        return root;
    }
};
var convertBST = function(root) {
    let sum = 0;
    const go = (node) => {
        if (!node) return;
        go(node.right);
        sum += node.val;
        node.val = sum;
        go(node.left);
    }
    go(root);
    return root;
};
Convert BST to Greater Tree