<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

# 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:
        child_to_par = defaultdict(int)
        par_to_child = defaultdict(list)
        visited = set([start])
        def dfs(root):
            if not root:
                return
            if root.left:
                par_to_child[root.val].append(root.left.val)
                child_to_par[root.left.val] = root.val
            if root.right:
                par_to_child[root.val].append(root.right.val)
                child_to_par[root.right.val] = root.val
            dfs(root.left)
            dfs(root.right)
        dfs(root)
        def bfs(start):
            queue = deque([start])
            paths = []
            while queue:
                level_size = len(queue)
                current_level = []  
                for _ in range(level_size):
                    node = queue.popleft()
                    current_level.append(node)                    
                    for child in par_to_child[node]:
                        if child not in visited:
                            visited.add(child)
                            queue.append(child)                    
                    if node in child_to_par and child_to_par[node] not in visited:
                        parent = child_to_par[node]
                        visited.add(parent)
                        queue.append(parent)                
                paths.append(current_level)
            return paths
        all_paths = bfs(start)
        print(all_paths)  
        return len(all_paths) - 1  # Assuming you want the "time" as the number of levels (path length)
502