<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 defaultdict, deque

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: TreeNode, start: int) -> int:
        # Step 1: Convert the tree to an undirected graph
        graph = defaultdict(list)
        
        def build_graph(node, parent=None):
            if not node:
                return
            if parent:
                graph[node.val].append(parent.val)
                graph[parent.val].append(node.val)
            build_graph(node.left, node)
            build_graph(node.right, node)
        
        build_graph(root)
        
        # Step 2: Perform BFS from the start node
        visited = set()
        queue = deque([(start, 0)])
        max_time = 0
        
        while queue:
            node, time = queue.popleft()
            if node in visited:
                continue
            visited.add(node)
            max_time = max(max_time, time)
            for neighbor in graph[node]:
                if neighbor not in visited:
                    queue.append((neighbor, time + 1))
        
        return max_time

# Helper function to build the tree for testing
def build_tree(nodes):
    if not nodes:
        return None
    root = TreeNode(nodes[0])
    queue = deque([root])
    i = 1
    while i < len(nodes):
        node = queue.popleft()
        if nodes[i] is not None:
            node.left = TreeNode(nodes[i])
            queue.append(node.left)
        i += 1
        if i < len(nodes) and nodes[i] is not None:
            node.right = TreeNode(nodes[i])
            queue.append(node.right)
        i += 1
    return root

# Example test cases
solution = Solution()

# Example 1
root1 = build_tree([1,5,3,None,4,10,6,9,2])
assert solution.amountOfTime(root1, 3) == 4

# Example 2
root2 = build_tree([1])
assert solution.amountOfTime(root2, 1) == 0

print("All test cases passed!")
428