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### by Sumit Chourasia | Oct 08, 2020 | Category :coding | Tags : algorithmbinary-treedata-structureeasyleetcodetree #### Trim a Binary Search Tree - Tree - Easy - LeetCode

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 = , low = 1, high = 2
Output: 
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: 

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

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     public int val;
*     public TreeNode left;
*     public TreeNode right;
*     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
public class Solution {
public TreeNode TrimBST(TreeNode root, int low, int high) {
if(root == null){
return null;
}

if(low>root.val){
return TrimBST(root.right,low,high);
}

if(high<root.val){
return TrimBST(root.left,low,high);
}

root.left = TrimBST(root.left,low,high);
root.right = TrimBST(root.right,low,high);

return root;
}
}``````

Time Complexity: O(n)

Space Complexity: O(1)