<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>

# 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 treeQueries(self, root: TreeNode, queries: List[int]) -> List[int]:
        from collections import defaultdict
        
        # To hold the depth and height of each node
        depth = {}
        height = {}
        
        # Calculate the depth of each node and the height of the subtree rooted at each node
        def dfs(node, d):
            if not node:
                return -1  # If the node is None, height is -1
            depth[node.val] = d
            left_height = dfs(node.left, d + 1)
            right_height = dfs(node.right, d + 1)
            node_height = 1 + max(left_height, right_height)
            height[node.val] = node_height
            return node_height
        
        # Precompute depth and height of each node
        dfs(root, 0)
        
        # For each depth level, find the two largest heights
        max_height_at_depth = defaultdict(list)
        def dfs_collect(node, d):
            if not node:
                return 
            # Maintain a list of two largest heights for each depth level
            if len(max_height_at_depth[d]) < 2:
                max_height_at_depth[d].append(height[node.val])
                max_height_at_depth[d].sort(reverse=True)
            else:
                if height[node.val] > max_height_at_depth[d][1]:
                    max_height_at_depth[d][1] = height[node.val]
                    max_height_at_depth[d].sort(reverse=True)
            dfs_collect(node.left, d + 1)
            dfs_collect(node.right, d + 1)
        
        # Collect heights for max_height_at_depth
        dfs_collect(root, 0)
        
        # Answer queries
        result = []
        for q in queries:
            d = depth[q]
            # Get the largest height not involving query node
            if len(max_height_at_depth[d]) == 1:
                # If there's only one entry at this depth, remove this depth entirely
                max_other_height = -1
            else:
                # Otherwise, take the second largest height
                if height[q] == max_height_at_depth[d][0]:
                    max_other_height = max_height_at_depth[d][1]
                else:
                    max_other_height = max_height_at_depth[d][0]
            
            # Compute the resulting height after subtree removal
            result.append(d + max_other_height)
        
        return result
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