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

Two Pointers

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

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//
// 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;
    }
};