<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>

def getHeight(n, hts):
    if not n:
        return 0

    hts[n] = max(getHeight(n.left, hts), getHeight(n.right, hts)) + 1
    return hts[n]

def populatePath(n, target, path):
    if n is None:
        return False, path

    path.append(n)
    if n.val == target:
        return True, path

    is_found, path = populatePath(n.left, target, path)
    if is_found: return True, path

    is_found, path = populatePath(n.right, target, path)
    if is_found: return True, path   

    path.pop()
    return False, path

class Solution:
    def amountOfTime(self, root: Optional[TreeNode], start: int) -> int:
        hts = {}
        getHeight(root, hts)
        # for n in hts: print(n.val, hts[n])

        path = []
        populatePath(root, start, path)

        max_burn_tm = hts[path[-1]]-1

        # print([n.val for n in path], max_burn_tm)
        tm = 1
        for i in range(len(path)-2, -1, -1):
            curr_burn_tm = tm
            if path[i].left == path[i+1] and path[i].right:
                # print("here", path[i].val)
                curr_burn_tm += hts[path[i].right]
            elif path[i].right == path[i+1] and path[i].left:
                    # print("here", path[i].val)
                    curr_burn_tm += hts[path[i].left]
            max_burn_tm = max(
                max_burn_tm,
                curr_burn_tm
            )
            # print(path[i].val, curr_burn_tm)
            tm += 1

        return max_burn_tm
        
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