<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) {
        
        std::map<int, std::vector<int>> nodes;

        auto dfs = [&nodes](decltype(root) root, auto&& dfs){

            if(root == nullptr) return;
            auto key_itr = nodes.find(root->val);
            if(key_itr == nodes.end())
                nodes[root->val] = {};
            if(root->left){
                nodes[root->val].push_back(root->left->val);
                dfs(root->left, dfs);
                nodes[root->left->val].push_back(root->val);
            }
            if(root->right){
                nodes[root->val].push_back(root->right->val);
                dfs(root->right, dfs);
                nodes[root->right->val].push_back(root->val);
            }
        };

        dfs(root, dfs);

        // for(auto [_key, _node_list]: nodes)
        // {
        //     std::cout << _key << ": ";
        //     for(auto e: _node_list)
        //     {
        //         std::cout << e << ", ";
        //     } 
        //     std::cout << '\n';
        // }

        std::set<int> visited;
        std::deque<int> q;
        q.push_back(start);
        int time = -1;
        while(q.size()!=0)
        {
            auto n = q.size();
            time++;
            while(n-- > 0)
            {
                auto node = q[0];
                q.pop_front();
                if(visited.find(node) != visited.end()) continue;
                for(auto _node: nodes[node])
                {
                    if(visited.find(_node) != visited.end()) continue;
                    q.push_back(_node);
                }
                visited.insert(node);
            }
        }

        return time;
    };

};


/*

2: 3
3: 2, 4, 1
4: 3 
1: 3

*/
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