<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: Optional[TreeNode], queries: List[int]) -> List[int]:
        # for each node, precompute dist_to_root
        # for each node, precompute dist_to_leaf (basically height)
        # query[i] = max(query[i.parent], dist_to_root[i.parent] + (dist_to_leaf[x.neighbor] if exists))

        # get size of tree and dist_to_root
        n = 0
        q = []
        q.append(root)
        idx_to_dist = {root.val: 0}
        while len(q) > 0:
            curr_node = q.pop(0)
            n += 1
            if curr_node.left is not None:
                idx_to_dist[curr_node.left.val] = idx_to_dist[curr_node.val] + 1
                q.append(curr_node.left)
            if curr_node.right is not None:
                idx_to_dist[curr_node.right.val] = idx_to_dist[curr_node.val] + 1
                q.append(curr_node.right)
        dist_to_root = [-1 for _ in range(n + 1)]
        for k, v in idx_to_dist.items():
            dist_to_root[k] = v

        # get parent and children graph 
        parent = [-1 for _ in range(n + 1)]
        childrens = [[] for _ in range(n + 1)]
        q = []
        q.append(root)
        orders = []
        while len(q) > 0:
            curr_node = q.pop(0)
            orders.append(curr_node.val)
            if curr_node.left is not None:
                parent[curr_node.left.val] = curr_node.val
                childrens[curr_node.val].append(curr_node.left.val)
                q.append(curr_node.left)
            if curr_node.right is not None:
                parent[curr_node.right.val] = curr_node.val
                childrens[curr_node.val].append(curr_node.right.val)
                q.append(curr_node.right)

        # compute dist_to_leaf in O(n) time
        dist_to_leaf = [-1 for _ in range(n + 1)]
        def get_dist_to_leaf(node):
            ans = 0
            if node.left is not None:
                ans = max(ans, 1 + get_dist_to_leaf(node.left))
            if node.right is not None:
                ans = max(ans, 1 + get_dist_to_leaf(node.right))
            dist_to_leaf[node.val] = ans
            return ans
        get_dist_to_leaf(root)

        query_ans = [-1 for _ in range(n + 1)]
        for val in orders:
            if parent[val] == -1:
                query_ans[val] = 0
                continue
            
            par = parent[val]
            temp = max(query_ans[par], dist_to_root[par])

            for child in childrens[par]:
                if child == val:
                    continue
                temp = max(temp, dist_to_root[par] + dist_to_leaf[child] + 1)
            
            query_ans[val] = temp

        #print(parent)
        #print(orders)
        #print(query_ans)

        ans = []
        for q in queries:
            ans.append(query_ans[q])

        return ans
        
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