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We can utilize the properties of a BST to perform a recursive traversal. The strategy here involves:
low, we need to trim the left subtree and consider the right subtree.high, we trim the right subtree and consider the left subtree.low, high], we recursively trim both subtrees.Time Complexity: O(n), where n is the number of nodes in the tree, since each node is processed once.
Space Complexity: O(h), where h is the height of the tree, representing the recursion stack.
1class TreeNode {
2 int val;
3 TreeNode left, right;
4 TreeNode(int x) { val = x; }
5}
6
7public class Solution {
8 public TreeNode trimBST(TreeNode root, int low, int high) {
9 if (root == null) return null;
10 if (root.val < low) return trimBST(root.right, low, high);
11 if (root.val > high) return trimBST(root.left, low, high);
12 root.left = trimBST(root.left, low, high);
13 root.right = trimBST(root.right, low, high);
14 return root;
15 }
16}The Java solution follows the same logic, leveraging recursion by returning results from left and right subtrees adjustments.
This iterative approach uses a stack to traverse the tree. The main idea is to mimic the recursive depth-first search using an explicit stack.
Time Complexity: O(n), as each node is processed once.
Space Complexity: O(h), where h is the height of the tree, due to the stack usage.
1import java.util.Stack;
This Java example shows an iterative approach using a stack to maintain the modified tree's structure while keeping nodes within bounds. Each node is processed, ensuring valid nodes are stacked for later traversal.