### Problem Statement

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:**

Problem linkInput:The root of a Binary Search Tree like this: 5 / \ 2 13Output:The root of a Greater Tree like this: 18 / \ 20 13

### Video Tutorial

You can find the detailed video tutorial here### Thought Process

When a tree problem is presented, normally we will think if any kind of traversal could be applied, e.g, preorder/inorder/postorder or level order traversal. In this problem, we want to update the tree node value with the sum of itself and all the node values larger itself. Yep, directly modifying the input param values (In real world, it is bad practice to directly manipulate the input value since it might surprise your downstream callers unless it is clearly documented), so tree structure is not modified.You might first think of a brute force method, perform a normal left to right inorder traversal, then it is sorted ascendingly, then do a rolling sum from largest to smallest element. This is normal when we don't really know how to deal with tree structure so we serialize the tree into an array and later we can deserialize it back. In this problem, we don't need to deserialize since tree structure is not changed. This however requires extra O(N) space.

Now let's look at a better solution. In BST, all the nodes that have greater value than the node itself lives in the right subtree, so if we traverse the right subtree first, update the tree value, then traverse the left subtree (all the right subtree nodes have greater value than left subtree), then the problem becomes doing an inorder traversal, except this time we go to right first, update value, then go to left. To keep the running sum, it is most easy to keep a global variable to track it.

### Solutions

#### Naive inorder traversal with serialize tree into array

Time Complexity: O(N), N is the total number of tree nodes

Space Complexity: O(N) , N is the total number of tree nodes, since we are using extra list.

#### Reverse inorder traversal

As described above, we simply do a reverse inorder traversal, right to left and keep a running sum.Time Complexity: O(N), N is the total number of tree nodes

Space Complexity: O(lgN) , N is the total number of tree nodes, since we are doing recursion.

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