[LeetCode February Challange] Day 2 - Trim a Binary Search Tree

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, 10^4].
• 0 <= Node.val <= 10^4
• The value of each node in the tree is unique.
• root is guaranteed to be a valid binary search tree.
• 0 <= low <= high <= 10^4

---

Solution

Time complexity : O(n)
Space complexity : O(n)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if (root == nullptr) return nullptr;
        if (root->val < low) 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;
    }
};