Closest binary search tree value ii

Two Pointers Stack Tree Depth-First Search Binary Search Tree Heap (Priority Queue) Binary Tree

61.2% acceptance

Mar 31, 2026

1343

Given the root of a binary search tree, a target value, and an integer k, return the k values in the BST that are closest to the target. You may return the answer in any order. You are guaranteed to have only one unique set of k values in the BST that are closest to the target.

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   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 closest_k_values(root: Option<Rc<RefCell<TreeNode>>>, target: f64, k: i32) -> Vec<i32> {
    use std::collections::VecDeque;
    let mut deque: VecDeque<i32> = VecDeque::new();
    Self::inorder(&root, target, k as usize, &mut deque);
    deque.into_iter().collect()
  }
  
  fn inorder(node: &Option<Rc<RefCell<TreeNode>>>, target: f64, k: usize, deque: &mut std::collections::VecDeque<i32>) {
    if let Some(n) = node {
      let n_ref = n.borrow();
      Self::inorder(&n_ref.left, target, k, deque);
      if deque.len() < k {
        deque.push_back(n_ref.val);
      } else if (n_ref.val as f64 - target).abs() < (*deque.front().unwrap() as f64 - target).abs() {
        deque.pop_front();
        deque.push_back(n_ref.val);
      } else {
        return;
      }
      Self::inorder(&n_ref.right, target, k, deque);
    }
  }
}