669. Trim a Binary Search Tree
Given the root
of a binary search tree and the lowest and highest boundaries as low
and high
, trim the tree so that all its elements lies in [low, high]
. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a unique answer.
Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.
Example 1:

Input: root = [1,0,2], low = 1, high = 2 Output: [1,null,2]
Example 2:

Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3 Output: [3,2,null,1]
Example 3:
Input: root = [1], low = 1, high = 2 Output: [1]
Example 4:
Input: root = [1,null,2], low = 1, high = 3 Output: [1,null,2]
Example 5:
Input: root = [1,null,2], low = 2, high = 4 Output: [2]
Constraints:
- The number of nodes in the tree in the range
[1, 104]
. 0 <= Node.val <= 104
- The value of each node in the tree is unique.
root
is guaranteed to be a valid binary search tree.0 <= low <= high <= 104
Rust Solution
struct Solution;
use rustgym_util::*;
impl Solution {
fn trim_bst(root: TreeLink, l: i32, r: i32) -> TreeLink {
if let Some(node) = root.clone() {
let mut node = node.borrow_mut();
let left = node.left.take();
let right = node.right.take();
if node.val > r {
return Self::trim_bst(left, l, r);
}
if node.val < l {
return Self::trim_bst(right, l, r);
}
node.left = Self::trim_bst(left, l, r);
node.right = Self::trim_bst(right, l, r);
root
} else {
None
}
}
}
#[test]
fn test() {
let root = tree!(1, tree!(0), tree!(2));
let res = tree!(1, None, tree!(2));
assert_eq!(Solution::trim_bst(root, 1, 2), res);
let root = tree!(3, tree!(0, None, tree!(2, tree!(1), None)), tree!(4));
let res = tree!(3, tree!(2, tree!(1), None), None);
assert_eq!(Solution::trim_bst(root, 1, 3), res);
}
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