<p>You are given the <code>root</code> of a <strong>binary search tree </strong>and an array <code>queries</code> of size <code>n</code> consisting of positive integers.</p>

<p>Find a <strong>2D</strong> array <code>answer</code> of size <code>n</code> where <code>answer[i] = [min<sub>i</sub>, max<sub>i</sub>]</code>:</p>

<ul>
	<li><code>min<sub>i</sub></code> is the <strong>largest</strong> value in the tree that is smaller than or equal to <code>queries[i]</code>. If a such value does not exist, add <code>-1</code> instead.</li>
	<li><code>max<sub>i</sub></code> is the <strong>smallest</strong> value in the tree that is greater than or equal to <code>queries[i]</code>. If a such value does not exist, add <code>-1</code> instead.</li>
</ul>

<p>Return <em>the array</em> <code>answer</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/28/bstreeedrawioo.png" style="width: 261px; height: 281px;" />
<pre>
<strong>Input:</strong> root = [6,2,13,1,4,9,15,null,null,null,null,null,null,14], queries = [2,5,16]
<strong>Output:</strong> [[2,2],[4,6],[15,-1]]
<strong>Explanation:</strong> We answer the queries in the following way:
- The largest number that is smaller or equal than 2 in the tree is 2, and the smallest number that is greater or equal than 2 is still 2. So the answer for the first query is [2,2].
- The largest number that is smaller or equal than 5 in the tree is 4, and the smallest number that is greater or equal than 5 is 6. So the answer for the second query is [4,6].
- The largest number that is smaller or equal than 16 in the tree is 15, and the smallest number that is greater or equal than 16 does not exist. So the answer for the third query is [15,-1].
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/28/bstttreee.png" style="width: 101px; height: 121px;" />
<pre>
<strong>Input:</strong> root = [4,null,9], queries = [3]
<strong>Output:</strong> [[-1,4]]
<strong>Explanation:</strong> The largest number that is smaller or equal to 3 in the tree does not exist, and the smallest number that is greater or equal to 3 is 4. So the answer for the query is [-1,4].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 10<sup>5</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>6</sup></code></li>
	<li><code>n == queries.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= queries[i] &lt;= 10<sup>6</sup></code></li>
</ul>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def __init__(self):
        self.inorder_vals = []  # This will store the inorder traversal values
    
    def inorder_traversal(self, root):
        """Perform inorder traversal to get a sorted list of values from the BST."""
        if root is None:
            return
        self.inorder_traversal(root.left)
        self.inorder_vals.append(root.val)
        self.inorder_traversal(root.right)
    
    def find_closest_smaller_or_equal(self, query):
        """Find the largest value smaller than or equal to the query using binary search."""
        low, high = 0, len(self.inorder_vals) - 1
        closest_smaller = -1
        
        while low <= high:
            mid = (low + high) // 2
            if self.inorder_vals[mid] > query:
                high = mid - 1
            else:
                closest_smaller = self.inorder_vals[mid]  # Potential candidate for the answer
                low = mid + 1
        
        return closest_smaller
    
    def find_closest_greater_or_equal(self, query):
        """Find the smallest value greater than or equal to the query using binary search."""
        low, high = 0, len(self.inorder_vals) - 1
        closest_greater = -1
        
        while low <= high:
            mid = (low + high) // 2
            if self.inorder_vals[mid] < query:
                low = mid + 1
            else:
                closest_greater = self.inorder_vals[mid]  # Potential candidate for the answer
                high = mid - 1
        
        return closest_greater
    
    def closestNodes(self, root: TreeNode, queries: list[int]) -> list[list[int]]:
        # Perform inorder traversal to get a sorted list of BST values
        self.inorder_traversal(root)
        
        result = []
        
        # For each query, find the closest smaller or equal and closest greater or equal values
        for query in queries:
            closest_smaller = self.find_closest_smaller_or_equal(query)
            closest_greater = self.find_closest_greater_or_equal(query)
            result.append([closest_smaller, closest_greater])
        
        return result
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