669. Trim a Binary Search Tree

1. Description

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.

2. Example

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]

3. 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 <= $10^4$

4. Solutions

My Accepted Solution

n is the number of nodes in m_root
Time complexity: O(n)
Space complexity: O(1)

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

class Solution 
{
public:
    // TreeNode* trimBST(TreeNode* root, int low, int high)
    TreeNode* trimBST(TreeNode *m_root, int low, int high) 
    {
        if(!m_root) return m_root;
        
        if(m_root->val < low) return trimBST(m_root->right, low, high);
        if(m_root->val > high) return trimBST(m_root->left, low, high);
        
        m_root->left = trimBST(m_root->left, low, high);
        m_root->right = trimBST(m_root->right, low, high);
        
        return m_root;
    }
};