<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.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int amountOfTime(TreeNode* root, int start) {
        
        unordered_map<TreeNode*,unordered_set<TreeNode*>>mp;

        queue<TreeNode*>qa;
        qa.push(root);

        TreeNode* s;

        while(qa.size()!=0)
        {
            int sz = qa.size();

            for(int i=0;i<sz;i++)
            {
                TreeNode* x = qa.front();
                qa.pop();

                if(x->val==start)
                {
                    s = x;
                }

                if(x->left!=NULL)
                {
                    mp[x].insert(x->left);
                    mp[x->left].insert(x);

                    qa.push(x->left);
                }

                if(x->right!=NULL)
                {
                    mp[x].insert(x->right);
                    mp[x->right].insert(x);

                    qa.push(x->right);
                }
            }
        }

        queue<TreeNode*>qb;
        qb.push(s);

        unordered_set<TreeNode*>vis;
        vis.insert(s);

        int curlvl=-1;

        while(qb.size()!=0)
        {
            curlvl++;

            int sz = qb.size();

            for(int i=0;i<sz;i++)
            {
                TreeNode* x = qb.front();
                qb.pop();

                for(auto adj:mp[x])
                {
                    if(!vis.contains(adj))
                    {
                        qb.push(adj);
                        vis.insert(adj);
                    }
                }
            }
        }

        return curlvl;
    }
};
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