Increasing order search tree

/*
 * Given the root of a binary search tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only one right child.
 * Example 1:
 * Input: root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
 * Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
 * Example 2:
 * Input: root = [5,1,7]
 * Output: [1,null,5,null,7]
 * Constraints:
 * The number of nodes in the given tree will be in the range [1, 100].
 * 0 <= Node.val <= 1000
 */

use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  fn inorder(node: &Option<Rc<RefCell<TreeNode>>>, result: &mut Vec<i32>) {
    if let Some(n) = node {
      let n = n.borrow();
      Self::inorder(&n.left, result);
      result.push(n.val);
      Self::inorder(&n.right, result);
    }
  }
  pub fn increasing_bst(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
    let mut vals = Vec::new();
    Self::inorder(&root, &mut vals);
    let dummy = Rc::new(RefCell::new(TreeNode::new(0)));
    let mut cur = dummy.clone();
    for v in vals {
      let next = Rc::new(RefCell::new(TreeNode::new(v)));
      cur.borrow_mut().right = Some(next.clone());
      cur = next;
    }
    dummy.borrow().right.clone()
  }
}