# [LeetCode] 669. Trim a Binary Search Tree

## Problem

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
``````

## Solution

### Recursive

``````class Solution {
public TreeNode trimBST(TreeNode root, int L, int R) {
if (root == null || L > R) return null;
if (root.val > R) return trimBST(root.left, L, R);
if (root.val < L) return trimBST(root.right, L, R);
if (root.val >= L && root.val <= R) {
root.left = trimBST(root.left, L, R);
root.right = trimBST(root.right, L, R);
}
return root;
}
}
``````

### Iterative

``````class Solution {
public TreeNode trimBST(TreeNode root, int L, int R) {
if (root == null) return null;
while (root.val < L || root.val > R) {
if (root.val < L) {
root = root.right;
}
if (root.val > R) {
root = root.left;
}
}
TreeNode dummy = root;
while (dummy != null) {
while (dummy.left != null && dummy.left.val < L) {
dummy.left = dummy.left.right;
}
dummy = dummy.left;
}

dummy = root;
while (dummy != null) {
while (dummy.right != null && dummy.right.val > R) {
dummy.right = dummy.right.left;
}
dummy = dummy.right;
}

return root;
}
}``````

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