<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 tree_dfs(self, node, depth, node_depth, node_height):
      if node is None:
          return -1
      node_depth[node.val] = depth
      height = max(self.tree_dfs(node.left, depth + 1, node_depth, node_height), self.tree_dfs(node.right, depth + 1, node_depth, node_height)) + 1
      node_height[node.val] = height
      
      return height
    def treeQueries(self, root: Optional[TreeNode], queries: List[int]) -> List[int]:
      result = []
      node_depth = {}
      node_height = {}
      self.tree_dfs(root, 0, node_depth, node_height)
      
      depth_level = defaultdict(list)
      for data in node_depth:
          depth = node_depth[data]
          if len(depth_level[depth]) < 2:
              depth_level[depth].append(data)
          else:
              h1 = node_height[depth_level[depth][0]]
              h2 = node_height[depth_level[depth][1]]
              
              # update 2 highest height nodes
              if min(h1, h2) < node_height[data]:
                  if h1 > h2:
                      depth_level[depth] = [depth_level[depth][0], data]
                  else:
                      depth_level[depth] = [depth_level[depth][1], data]
      
      for query in queries:
          depth = node_depth[query]
          sibs = depth_level[depth][:]
          if query not in sibs: # removed node not in 2 highest height nodes
              max_height = max(node_height[sibs[0]], node_height[sibs[1]])
              result.append(depth + max_height)
          else:
              sibs.remove(query)
              if len(sibs) == 0: # this depth only has removed node
                  result.append(depth - 1)
              else:
                  remain = sibs[0]
                  result.append(depth + node_height[remain])
              
      return result
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