<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>

from collections import deque

# 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 amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
        # Step 1: Build the parent map and find the start node
        parent_map = {}
        start_node = None
        
        def dfs(node, parent):
            if not node:
                return
            nonlocal start_node
            if node.val == start:
                start_node = node
            parent_map[node] = parent
            dfs(node.left, node)
            dfs(node.right, node)
        
        dfs(root, None)
        
        # Step 2: Perform BFS from the start node
        visited = set()
        queue = deque([(start_node, 0)])
        max_minutes = 0
        
        while queue:
            node, minutes = queue.popleft()
            if node in visited:
                continue
            visited.add(node)
            max_minutes = max(max_minutes, minutes)
            
            # Add children and parent to the queue
            if node.left:
                queue.append((node.left, minutes + 1))
            if node.right:
                queue.append((node.right, minutes + 1))
            if parent_map.get(node):
                queue.append((parent_map[node], minutes + 1))
        
        # Step 3: Return the result
        return max_minutes

# Example usage:
# root = TreeNode(1)
# root.left = TreeNode(5)
# root.right = TreeNode(3)
# root.left.right = TreeNode(4)
# root.right.left = TreeNode(10)
# root.right.right = TreeNode(6)
# start = 3
# sol = Solution()
# print(sol.amountOfTime(root, start))  # Output: 4
58772