**Intuition**

The problem involves calculating the height of a binary tree after removing subtrees rooted at specific nodes. To efficiently handle multiple queries, we can preprocess the tree to store information about node depths and levels. By leveraging this information, we can quickly determine the new height of the tree after each removal.

**Approach**

**Preprocess the Tree:**- Perform a depth-first search (DFS) to calculate the depth and level of each node.
- Store this information in a
`nodeToDepthAndLevel`

map. - Maintain a
`levelToMaxDepths`

map to store the two largest depths for each level.

**Process Queries:**- For each query:
- Retrieve the depth and level of the node to be removed.
- Consult the
`levelToMaxDepths`

map to find the two largest depths remaining at that level. - Calculate the new height of the tree based on the remaining maximum depths.

- For each query:

**Complexity**

**Time complexity:**O(N + Q), where N is the number of nodes in the tree and Q is the number of queries. The preprocessing DFS takes O(N) time, and each query can be processed in O(1) time.**Space complexity:**O(N), primarily due to the`nodeToDepthAndLevel`

and`levelToMaxDepths`

maps.

## Problem

You are given the `root`

of a **binary tree** with `n`

nodes. Each node is assigned a unique value from `1`

to `n`

. You are also given an array `queries`

of size `m`

.

You have to perform `m`

**independent** queries on the tree where in the `i`

query you do the following:^{th}

**Remove**the subtree rooted at the node with the value`queries[i]`

from the tree. It is**guaranteed**that`queries[i]`

will**not**be equal to the value of the root.

Return *an array *`answer`

* of size *`m`

* where *`answer[i]`

* is the height of the tree after performing the *`i`

^{th}* query*.

**Note**:

- The queries are independent, so the tree returns to its
**initial**state after each query. - The height of a tree is the
**number of edges in the longest simple path**from the root to some node in the tree.

**Example 1:**

Input:root = [1,3,4,2,null,6,5,null,null,null,null,null,7], queries = [4]Output:[2]Explanation: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 -> 3 -> 2).

**Example 2:**

Input:root = [5,8,9,2,1,3,7,4,6], queries = [3,2,4,8]Output:[3,2,3,2]Explanation:We have the following queries: - Removing the subtree rooted at node with value 3. The height of the tree becomes 3 (The path 5 -> 8 -> 2 -> 4). - Removing the subtree rooted at node with value 2. The height of the tree becomes 2 (The path 5 -> 8 -> 1). - Removing the subtree rooted at node with value 4. The height of the tree becomes 3 (The path 5 -> 8 -> 2 -> 6). - Removing the subtree rooted at node with value 8. The height of the tree becomes 2 (The path 5 -> 9 -> 3).

**Constraints:**

- The number of nodes in the tree is
`n`

. `2 <= n <= 10`

^{5}`1 <= Node.val <= n`

- All the values in the tree are
**unique**. `m == queries.length`

`1 <= m <= min(n, 10`

^{4})`1 <= queries[i] <= n`

`queries[i] != root.val`

**Code**

PHP

```
class Solution {
private $nodeToDepthAndLevel = [];
private $levelToMaxDepths = [];
public function treeQueries($root, $queries) {
$this->traverse($root, 0);
$result = [];
foreach ($queries as $query) {
[$depth, $level] = $this->nodeToDepthAndLevel[$query] ?? [0, 0];
$maxDepths = $this->levelToMaxDepths[$level] ?? [0, 0];
$result[] = $level - 1 + max($maxDepths[0], $maxDepths[1] < $depth ? $maxDepths[1] : 0);
}
return $result;
}
private function traverse($node, $level) {
if (!$node) {
return 0;
}
$depth = max($this->traverse($node->left, $level + 1), $this->traverse($node->right, $level + 1)) + 1;
$this->nodeToDepthAndLevel[$node->val] = [$depth, $level];
$maxDepths = $this->levelToMaxDepths[$level] ?? [0, 0];
if ($depth > $maxDepths[1]) {
$maxDepths[0] = $maxDepths[1];
$maxDepths[1] = $depth;
} elseif ($depth > $maxDepths[0]) {
$maxDepths[0] = $depth;
}
$this->levelToMaxDepths[$level] = $maxDepths;
return $depth;
}
}
```