<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>

from collections import defaultdict

class Solution:
    def treeQueries(self, root: Optional[TreeNode], queries: List[int]) -> List[int]:
        # Store depth and height information of each node
        depth_height = defaultdict(lambda: [0, 0])
        max_height_at_depth = defaultdict(list)  # To store the two largest heights at each depth
        
        # Perform DFS to compute the depth and height of each node
        def DFS(node, depth):
            if node is None:
                return 0
            # Recur for left and right children to compute their heights
            left_height = DFS(node.left, depth + 1)
            right_height = DFS(node.right, depth + 1)
            
            # Compute the height of the current node
            node_height = max(left_height, right_height) + 1
            
            # Store the depth and height for the current node
            depth_height[node.val] = [depth, node_height]
            
            # Store the height for this depth
            max_height_at_depth[depth].append(node_height)
            max_height_at_depth[depth].sort(reverse=True)
            if len(max_height_at_depth[depth]) > 2:
                max_height_at_depth[depth] = max_height_at_depth[depth][:2]
            
            return node_height
        
        # Step 1: Populate depth_height and max_height_at_depth
        DFS(root, 0)
        
        # The full tree height before any queries (height of the root node)
        full_tree_height = depth_height[root.val][1]
        
        # Prepare the answer list
        result = []
        
        # Step 2: Process each query
        for q in queries:
            node_depth, node_height = depth_height[q]
            
            # If the node to be removed has the largest height at its depth,
            # we need to check the second largest height at that depth to adjust the tree height.
            if max_height_at_depth[node_depth][0] == node_height:
                # If the removed node has the largest height, use the second largest height
                if len(max_height_at_depth[node_depth]) > 1:
                    new_height = full_tree_height - node_height + max_height_at_depth[node_depth][1]
                else:
                    # If there's no second largest height, reduce the tree height accordingly
                    new_height = full_tree_height - node_height
            else:
                # If the removed node doesn't affect the largest height, tree height remains the same
                new_height = full_tree_height
            
            # Append the result for this query
            result.append(new_height - 1)
        
        return result

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