<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>

# # 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:
#         N = 10**5+1
#         visited = [False]*N
#         pa = defaultdict(TreeNode)
#         q = []
#         def dfs(node):
#             if not node:
#                 return
            
#             l = node.left
#             r = node.right
            
#             if l:
#                 pa[l] = node
#                 if l.val == start:
#                     if l.left:
#                         q.append(l.left)
#                         visited[l.left.val] = True
#                     if l.right:
#                         q.append(l.right)
#                         visited[l.right.val] = True
#                     q.append(node)
#                     visited[node.val] = True
#                 dfs(l)
#             if r:
#                 pa[r] = node
#                 if r.val == start:
#                     if r.left:
#                         q.append(r.left)
#                         visited[r.left.val] = True
#                     if r.right:
#                         q.append(r.right)
#                         visited[r.right.val] = True
#                     q.append(node)
#                     visited[node.val] = True
#                 dfs(r)
#         dfs(root)
        
#         print(f"q = {q}")
#         res = 0
#         while q:
#             lq = len(q)
#             while lq:
#                 curr = q.pop(0)
#                 if not visited[pa[curr].val]:
#                     q.append(pa[curr])
#                     visited[pa[curr].val] = True
#                 if curr.left and not visited[curr.left.val]:
#                     q.append(curr.left)
#                     visited[curr.left.val] = True
#                 if curr.right and not visited[curr.right.val]:
#                     q.append(curr.right)
#                     visited[curr.right.val] = True
#                 lq -= 1
#             res += 1
#         return res
from collections import defaultdict, deque

class Solution:
    def amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
        pa = {}
        q = deque()
        visited = set()
        
        def dfs(node, parent):
            if not node:
                return
            pa[node.val] = parent
            if node.val == start:
                q.append(node)
                visited.add(node.val)
            dfs(node.left, node)
            dfs(node.right, node)
        
        dfs(root, None)
        
        res = 0
        
        while q:
            for _ in range(len(q)):
                curr = q.popleft()
                # Check the parent
                if curr.val in pa and pa[curr.val] and pa[curr.val].val not in visited:
                    q.append(pa[curr.val])
                    visited.add(pa[curr.val].val)
                # Check the left child
                if curr.left and curr.left.val not in visited:
                    q.append(curr.left)
                    visited.add(curr.left.val)
                # Check the right child
                if curr.right and curr.right.val not in visited:
                    q.append(curr.right)
                    visited.add(curr.right.val)
            res += 1
        
        return res - 1  # Subtract 1 to correct the off-by-one error

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