<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 setup_parents_and_find_start_node(self, root, parent, start):
        if not root: return
        # Better to use hashmap, instead of modifying the nodes
        root.parent = parent
        if root.val == start:
            self.source = root
        self.setup_parents_and_find_start_node(root.left, root, start)
        self.setup_parents_and_find_start_node(root.right, root, start)

    def amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
        # setup parent pointers to be able to traverse the tree like a Graph
        self.setup_parents_and_find_start_node(root, None, start)
        # Do a BFS and count the minutes

        queue = deque()
        # s = self.find_source(root, start)
        queue.append(self.source)
        infected = set()
        minutes = 0

        # using BFS
        # while queue:
        #     f = False
        #     for _ in range(len(queue)):
        #         front = queue.popleft()
        #         if front.val in infected: continue
        #         infected.add(front.val)
        #         for adj in ( front.parent, front.left, front.right):
        #             if adj and adj.val not in infected:
        #                 f = True
        #                 queue.append(adj)
        #     if f: minutes += 1
        # return minutes

        # using DFS
        return self.find_the_farthest_node(self.source, infected) - 1
    
    def find_the_farthest_node(self, start_node, visited):
        if not start_node: return 0
        if start_node in visited: return 0
        visited.add(start_node)
        u = self.find_the_farthest_node(start_node.left, visited)
        l = self.find_the_farthest_node(start_node.right, visited)
        r = self.find_the_farthest_node(start_node.parent, visited)
        return 1 + max(l, r, u)
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