Construct binary tree from preorder and postorder traversal

/*
 * Given two integer arrays, preorder and postorder where preorder is the preorder traversal of a binary tree of distinct values and postorder is the postorder traversal of the same tree, reconstruct and return the binary tree.
 * If there exist multiple answers, you can return any of them.
 * Example 1:
 * Input: preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
 * Output: [1,2,3,4,5,6,7]
 * Example 2:
 * Input: preorder = [1], postorder = [1]
 * Output: [1]
 * Constraints:
 * 1 <= preorder.length <= 30
 * 1 <= preorder[i] <= preorder.length
 * All the values of preorder are unique.
 * postorder.length == preorder.length
 * 1 <= postorder[i] <= postorder.length
 * All the values of postorder are unique.
 * It is guaranteed that preorder and postorder are the preorder traversal and postorder traversal of the same binary tree.
 */

use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  pub fn construct_from_pre_post(preorder: Vec<i32>, postorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
    fn build(pre: &[i32], post: &[i32]) -> Option<Rc<RefCell<TreeNode>>> {
      if pre.is_empty() { return None; }
      let root = Rc::new(RefCell::new(TreeNode::new(pre[0])));
      if pre.len() == 1 { return Some(root); }
      let l_val = pre[1];
      let l_size = post.iter().position(|&x| x == l_val).unwrap() + 1;
      root.borrow_mut().left = build(&pre[1..1+l_size], &post[..l_size]);
      root.borrow_mut().right = build(&pre[1+l_size..], &post[l_size..post.len()-1]);
      Some(root)
    }
    build(&preorder, &postorder)
  }
}