<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>

class Solution {
public:
    void traverse(TreeNode* node, int & max_nodes){
        if(node == NULL) return;

        max_nodes = max(max_nodes, node -> val);
        if(node -> left) traverse(node -> left, max_nodes);
        if(node -> right) traverse(node -> right, max_nodes);
    }
    int amountOfTime(TreeNode* root, int start) {
        queue<TreeNode*> q; q.push(root);

        int max_nodes = 0;
        traverse(root, max_nodes);
        vector<int> adj[max_nodes+1]; 
        
        while(!q.empty()){
            TreeNode* node = q.front(); q.pop();
            if(node -> left) {
                q.push(node -> left);
                adj[node -> val].push_back(node -> left -> val);
                adj[node -> left -> val].push_back(node -> val);
            }
            if(node -> right) {
                q.push(node -> right);
                adj[node -> val].push_back(node -> right -> val);
                adj[node -> right -> val].push_back(node -> val);
            }
        }   
        int count = 0;
        for(auto it : adj){
            cout << count << " - " ; count ++;
            for(auto kt : it){
                cout << kt << " ";
            }cout << endl;
        }

        vector<int> vis(max_nodes+1, 0); vis[start] = 1;
        queue<pair<int, int>> q2;
        q2.push({start, 0});
        int total_time = 0;
        while(!q2.empty()){
            int node = q2.front().first, time = q2.front().second; q2.pop();
            total_time = max(total_time, time);
            for(int k = 0; k < adj[node].size(); k++){
                if(!vis[adj[node][k]]){
                    vis[adj[node][k]] = 1;
                    q2.push({adj[node][k], time+1});
                }
            }
        }
        return total_time; 
    }
};
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