<p>You are given the <code>root</code> of a <strong>binary tree</strong> with <code>n</code> nodes. Each node is assigned a unique value from <code>1</code> to <code>n</code>. You are also given an array <code>queries</code> of size <code>m</code>.</p>

<p>You have to perform <code>m</code> <strong>independent</strong> queries on the tree where in the <code>i<sup>th</sup></code> query you do the following:</p>

<ul>
	<li><strong>Remove</strong> the subtree rooted at the node with the value <code>queries[i]</code> from the tree. It is <strong>guaranteed</strong> that <code>queries[i]</code> will <strong>not</strong> be equal to the value of the root.</li>
</ul>

<p>Return <em>an array </em><code>answer</code><em> of size </em><code>m</code><em> where </em><code>answer[i]</code><em> is the height of the tree after performing the </em><code>i<sup>th</sup></code><em> query</em>.</p>

<p><strong>Note</strong>:</p>

<ul>
	<li>The queries are independent, so the tree returns to its <strong>initial</strong> state after each query.</li>
	<li>The height of a tree is the <strong>number of edges in the longest simple path</strong> from the root to some node in the tree.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/07/binaryytreeedrawio-1.png" style="width: 495px; height: 281px;" />
<pre>
<strong>Input:</strong> root = [1,3,4,2,null,6,5,null,null,null,null,null,7], queries = [4]
<strong>Output:</strong> [2]
<strong>Explanation:</strong> The diagram above shows the tree after removing the subtree rooted at node with value 4.
The height of the tree is 2 (The path 1 -&gt; 3 -&gt; 2).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/07/binaryytreeedrawio-2.png" style="width: 301px; height: 284px;" />
<pre>
<strong>Input:</strong> root = [5,8,9,2,1,3,7,4,6], queries = [3,2,4,8]
<strong>Output:</strong> [3,2,3,2]
<strong>Explanation:</strong> We have the following queries:
- Removing the subtree rooted at node with value 3. The height of the tree becomes 3 (The path 5 -&gt; 8 -&gt; 2 -&gt; 4).
- Removing the subtree rooted at node with value 2. The height of the tree becomes 2 (The path 5 -&gt; 8 -&gt; 1).
- Removing the subtree rooted at node with value 4. The height of the tree becomes 3 (The path 5 -&gt; 8 -&gt; 2 -&gt; 6).
- Removing the subtree rooted at node with value 8. The height of the tree becomes 2 (The path 5 -&gt; 9 -&gt; 3).
</pre>

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

<ul>
	<li>The number of nodes in the tree is <code>n</code>.</li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= Node.val &lt;= n</code></li>
	<li>All the values in the tree are <strong>unique</strong>.</li>
	<li><code>m == queries.length</code></li>
	<li><code>1 &lt;= m &lt;= min(n, 10<sup>4</sup>)</code></li>
	<li><code>1 &lt;= queries[i] &lt;= n</code></li>
	<li><code>queries[i] != root.val</code></li>
</ul>

class Solution:
    def treeQueries(self, root: Optional[TreeNode], queries: List[int]) -> List[int]:
        
        # If the root is None, return an empty list
        if not root:
            return []

        # Dictionary to store the height of each node
        height = {}

        # Function to calculate and store the height of each node in the tree
        def set_heights(node):
            if not node:
                return -1
            # Calculate the height of the current node
            node_height = max(set_heights(node.left), set_heights(node.right)) + 1
            # Store the height of the current node in the height dictionary
            height[node.val] = node_height
            return node_height

        # Calculate and store heights starting from the root
        set_heights(root)

        # Dictionary to cache the second best height without the queried node
        cache = {}

        # Function to compute the second best height without a particular node
        def set_the_second_best_height_if_not_node(
            node, second_best_height_without_node, depth
        ):
            if not node:
                return

            # Cache the second best height without the current node
            cache[node.val] = second_best_height_without_node
            
            # Get the height of the left and right child nodes
            left_node_height = height[node.left.val] if node.left else -1
            right_node_height = height[node.right.val] if node.right else -1

            # Determine the maximum height without the current node
            if left_node_height > right_node_height:
                second_best_height_without_node = max(
                    second_best_height_without_node, right_node_height + depth + 1
                )
                # Recursively compute the second best height without the left child node
                set_the_second_best_height_if_not_node(
                    node.left, second_best_height_without_node, depth + 1
                )

            else:
                second_best_height_without_node = max(
                    second_best_height_without_node, left_node_height + depth + 1
                )
                # Recursively compute the second best height without the right child node
                set_the_second_best_height_if_not_node(
                    node.right, second_best_height_without_node, depth + 1
                )

        # Compute the second best height without each node
        set_the_second_best_height_if_not_node(root, -1, 0)

        # Process queries
        result = []
        for query in queries:
            # If the queried node is not in the cache, append the height of the root
            if query not in cache:
                result.append(height[root.val])
            # Otherwise, append the second best height without the queried node
            else:
                result.append(cache[query])

        return result
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