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
/**
* 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)
