<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:
        d= dict()

        def dfs(node, parent):
            if not node:
                return 

            if parent:
                d[node.val] = [parent.val]
            else:
                d[node.val] = []

            if node.left:
                d[node.val].append(node.left.val)
                dfs(node.left, node)

            if node.right:
                d[node.val].append(node.right.val)
                dfs(node.right, node)

        dfs(root, None)

        print(d)

        visit = set()
        stack = [start]
        visit.add(start)
        time = 0

        while stack:
            new_stack = []
            for nodeVal in stack:
                for v in d[nodeVal]:
                    if v not in visit:
                        visit.add(v)
                        new_stack.append(v)
            stack = new_stack
            time += 1

        return time - 1




        # d = dict()

        # def dfs(node, parent):
        #     if not node:
        #         return
        #     d[node.val] = set()
        #     if parent:
        #         d[node.val].add(parent.val)
        #     if node.left:
        #         d[node.val].add(node.left.val)
        #         dfs(node.left, node)
        #     if node.right:
        #         d[node.val].add(node.right.val)
        #         dfs(node.right, node)

        # dfs(root, None)
        # # print(d)

        # stack = [start]
        # count = 0
        # visited = set()
        # while stack:
        #     new_stack = []
        #     for node in stack:
        #         if node not in visited:
        #             visited.add(node)
        #             for i in d[node]:
        #                 if i not in visited: new_stack.append(i)
        #     stack = new_stack
        #     # print(stack)
        #     count += 1
        # return count - 1










        # d = dict()
        
        # def dfs(node, parent):
        #     if node:
        #         d[node.val] = []
        #         if parent:
        #             d[node.val].append(parent.val)
        #         if node.left:
        #             d[node.val].append(node.left.val)
        #             dfs(node.left, node)
        #         if node.right:
        #             d[node.val].append(node.right.val)
        #             dfs(node.right, node)
        # dfs(root, None)
        # # print(d)
        
        # stack = d[start]
        # mins = 0
        # visited = {start}
        # # print(mins, stack, visited)
        
        
        # while stack:
        #     new_stack = []
        #     # print(mins, stack, visited)
        #     for node in stack:
        #         if node not in visited:
        #             visited.add(node)
        #             for n in d[node]:
        #                 if n not in visited:
        #                     new_stack.append(n)
                        
        #     mins += 1
        #     # print(mins, stack, visited)
        #     stack = new_stack
            
        # return mins


                

        
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