538. Convert BST to Greater Tree

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 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.

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

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]
```

Example 3:

```Input: root = [1,0,2]
Output: [3,3,2]
```

Example 4:

```Input: root = [3,2,4,1]
Output: [7,9,4,10]
```

Constraints:

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

Rust Solution

``````struct Solution;
use rustgym_util::*;

trait Inorder {
fn inorder(&mut self, sum: &mut i32);
}

fn inorder(&mut self, sum: &mut i32) {
if let Some(node) = self {
let mut node = node.borrow_mut();
node.right.inorder(sum);
*sum += node.val;
node.val = *sum;
node.left.inorder(sum);
}
}
}

impl Solution {
let mut sum = 0;
root.inorder(&mut sum);
root
}
}

#[test]
fn test() {
let root = tree!(5, tree!(2), tree!(13));
let greater = tree!(18, tree!(20), tree!(13));
assert_eq!(Solution::convert_bst(root), greater);
}
``````

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