<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
"""

First step:
- determine height of each node

Second step:
- locate level, at which we are removing
- find other nodes on that level, and take max height of those nodes,
- then add depth+height to get resultant height of tree

- dfs: determine height of each node
- bfs: determine which nodes are on what level and save level per node
"""


class Solution:
    
    #dfs tree to determine max height of each node
    def dfs(self, node, height):
        
        # base case, leaf node, height is 0, return and save value in maps
        if not node.right and not node.left:
            self.node_to_level[node.val] = height
            self.level_heights[height].append(0)
            return 0
        
        # get the height of the left and right subtrees
        left_max = self.dfs(node.left, height+1) if node.left else -inf
        right_max = self.dfs(node.right, height+1) if node.right else -inf
        
        # get the max height
        max_height = max(left_max, right_max)
        
        # save the current nodes level and append the max_height to the heights on this level
        self.node_to_level[node.val] = height
        self.level_heights[height].append(max_height+1)
        self.node_to_height[node.val] = max_height+1
        
        return max_height+1
        
            
    
    def treeQueries(self, root: Optional[TreeNode], queries: List[int]) -> List[int]:
        
        self.node_to_level = defaultdict(int) # hold the level to which each node belongs to
        self.level_heights = defaultdict(list) # holds all height of all subtrees at the given level
        self.node_to_height = defaultdict(int) # we need this mapping because we need to check if we are removing a node with the greatest height
        
        answer = [] # store answers of our queries
        
        #load our maps with a dfs
        self.dfs(root, 0)
        
        # sort each list in level_heights in ascending order of height
        for level in self.level_heights.keys():
            self.level_heights[level].sort()
        
        # loop through queries
        for q in queries:
            
            #obtain level of node to remove
            level = self.node_to_level[q]
            
            # check if this level only has one node, if so we just deduct 1 from current height
            if len(self.level_heights[level])==1:
                answer.append(level-1)
            else:
                
                if self.level_heights[level][-1]==self.node_to_height[q]:
                    answer.append(self.level_heights[level][-2]+level)
                else:
                    answer.append(self.level_heights[level][-1]+level)
        
        return answer
                

        
        
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