# Trim a Binary Search Tree

Updated: Mar 24, 2021

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

**Solution:**

**Iterative:**

```
public TreeNode trimBST(TreeNode root, int low, int high) {
if (root == null) {
return null;
}
TreeNode dummyHead = new TreeNode();
Stack<List<TreeNode>> stack = new Stack<>();
stack.push(Arrays.asList(root, dummyHead));
TreeNode currentNode;
List<TreeNode> pair;
while (!stack.isEmpty()) {
pair = stack.pop();
currentNode = pair.get(0);
if (currentNode.val >= high) {
currentNode.right = null;
}
if (currentNode.val <= low) {
currentNode.left = null;
}
if (currentNode.val > low && currentNode.left != null) {
stack.push(Arrays.asList(currentNode.left, currentNode.val > high ? pair.get(1) : currentNode));
}
if (currentNode.val < high && currentNode.right != null) {
stack.push(Arrays.asList(currentNode.right, currentNode.val < low ? pair.get(1) : currentNode));
}
if (pair.get(1) == dummyHead || currentNode.val < pair.get(1).val) {
pair.get(1).left = currentNode;
} else {
pair.get(1).right = currentNode;
}
}
return dummyHead.left;
}
```

**Recursive:**

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