<p>You are given the <code>root</code> of a binary tree with <strong>unique</strong> values, and an integer <code>start</code>. At minute <code>0</code>, an <strong>infection</strong> starts from the node with value <code>start</code>.</p>

<p>Each minute, a node becomes infected if:</p>

<ul>
	<li>The node is currently uninfected.</li>
	<li>The node is adjacent to an infected node.</li>
</ul>

<p>Return <em>the number of minutes needed for the entire tree to be infected.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/25/image-20220625231744-1.png" style="width: 400px; height: 306px;" />
<pre>
<strong>Input:</strong> root = [1,5,3,null,4,10,6,9,2], start = 3
<strong>Output:</strong> 4
<strong>Explanation:</strong> The following nodes are infected during:
- Minute 0: Node 3
- Minute 1: Nodes 1, 10 and 6
- Minute 2: Node 5
- Minute 3: Node 4
- Minute 4: Nodes 9 and 2
It takes 4 minutes for the whole tree to be infected so we return 4.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/06/25/image-20220625231812-2.png" style="width: 75px; height: 66px;" />
<pre>
<strong>Input:</strong> root = [1], start = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> At minute 0, the only node in the tree is infected so we return 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>5</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
	<li>Each node has a <strong>unique</strong> value.</li>
	<li>A node with a value of <code>start</code> exists in the tree.</li>
</ul>

class Solution:
    def amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
        # lets first locate start node
        parent = {}

        def find_start(node, start, parent):
            if not node:
                return
            if node.val == start:
                return node
            parent[node.left] = node
            parent[node.right] = node
            return find_start(node.left, start, parent) or find_start(node.right, start, parent)

        start_node = find_start(root, start, parent)

        # now we have the start node, we can do a bfs to find the time

        def dfs(node, parent):
            # q = deque([(node, 0)]) 
            q = [(node, 0)]
            visited = set()
            res = 0

            while q:
                node, time = q.pop()
                if node in visited:
                    continue

                res = max(res, time)    
                visited.add(node)
                if node.left and node.left not in visited:
                    q.append((node.left, time + 1))
                if node.right and node.right not in visited:
                    q.append((node.right, time + 1))
                if node in parent and parent[node] not in visited:
                    q.append((parent[node], time + 1))
            
            return res

        return dfs(start_node, parent)
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