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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.
1#include <stdio.h>
2#include <stdlib.h>
3
4struct TreeNode {
5 int val;
6 struct TreeNode *left;
7 struct TreeNode *right;
8};
9
10struct TreeNode* trimBST(struct TreeNode* root, int low, int high) {
11 if (!root) return NULL;
12 if (root->val < low) return trimBST(root->right, low, high);
13 if (root->val > high) return trimBST(root->left, low, high);
14 root->left = trimBST(root->left, low, high);
15 root->right = trimBST(root->right, low, high);
16 return root;
17}
The recursive function trimBST
checks if the node is NULL. If the node's value is less than low
, it trims the left subtree. If the node's value is greater than high
, it trims the right subtree. Otherwise, it recursively processes both subtrees.
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.
1class TreeNode:
2
The Python solution employs a stack for an iterative node traversal. Adjustments are made according to the node values relative to low
and high
. Nodes are appended to a stack for traversal if they are valid.