Given a binary search tree and the lowest and highest boundaries as L
and R
,
trim the tree so that all its elements lies in [L, R]
(R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.
Example 1:
Input:
1
/ \
0 2
L = 1
R = 2
Output:
1
\
2
Example 2:
Input:
3
/ \
0 4
\
2
/
1
L = 1
R = 3
Output:
3
/
2
/
1
O(log[base 2](n)) Time - Recursive:
/**
* Definition for a binary tree node.
* public class TreeNode {
* public var val: Int
* public var left: TreeNode?
* public var right: TreeNode?
* public init(_ val: Int) {
* self.val = val
* self.left = nil
* self.right = nil
* }
* }
*/
class Solution {
func trimBST(_ root: TreeNode?, _ L: Int, _ R: Int) -> TreeNode? {
guard let node = root else { return nil }
switch node.val {
case Int.min..<L:
return trimBST(node.right, L, R)
case R+1...Int.max:
return trimBST(node.left, L, R)
default:
node.left = trimBST(node.left, L, R)
node.right = trimBST(node.right, L, R)
return node
}
}
}