/*
 * @lc app=leetcode id=938 lang=rust
 *
 * [938] Range Sum of BST
 *
 * https://leetcode.com/problems/range-sum-of-bst/description/
 *
 * algorithms
 * Easy (79.26%)
 * Total Accepted:    152K
 * Total Submissions: 191.8K
 * Testcase Example:  '[10,5,15,3,7,null,18]\n7\n15'
 *
 * Given the root node of a binary search tree, return the sum of values of all
 * nodes with value between L and R (inclusive).
 * 
 * The binary search tree is guaranteed to have unique values.
 * 
 * 
 * 
 * 
 * Example 1:
 * 
 * 
 * Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
 * Output: 32
 * 
 * 
 * 
 * Example 2:
 * 
 * 
 * Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
 * Output: 23
 * 
 * 
 * 
 * 
 * Note:
 * 
 * 
 * The number of nodes in the tree is at most 10000.
 * The final answer is guaranteed to be less than 2^31.
 * 
 * 
 * 
 */
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// // pub structTreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }

// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }

use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    pub fn range_sum_bst(root: Option<Rc<RefCell<TreeNode>>>, l: i32, r: i32) -> i32 {
        if let Some(root) = root {
            let mut _sum = 0; 
            if l <= root.borrow().val && r >= root.borrow().val { _sum += root.borrow().val ;}
            if l < root.borrow().val { _sum += Self::range_sum_bst(root.borrow().left.clone(), l, r); }
            if r > root.borrow().val { _sum += Self::range_sum_bst(root.borrow().right.clone(), l, r); }
           return _sum ;
        }
        0
    }
}

// pub structSolution;