<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:
    map<int, vector<int>> adjList;
    void inorder(TreeNode* root){
        if(root == NULL){
            return;
        }
        if(root->left != NULL){
            adjList[root->val].push_back(root->left->val);
            adjList[root->left->val].push_back(root->val);
            inorder(root->left);
        }
        if(root->right != NULL){
            adjList[root->val].push_back(root->right->val);
            adjList[root->right->val].push_back(root->val);
            inorder(root->right);
        }
    }
    int amountOfTime(TreeNode* root, int start) {
        inorder(root);
        /*for(auto itr = adjList.begin(); itr!=adjList.end();itr++){
            int u = itr->first;
            cout<<u<<": "<<" ";
            for(int v : itr->second){
                cout<<v<<" ";
            }
            cout<<endl;
        }*/
        queue<int> q;
        set<int> visited;
        visited.insert(start);
        for(int v : adjList[start]){
            if(visited.find(v) == visited.end()){
                q.push(v);
            }
        }
        int time = 0;
        while(!q.empty()){
            //cout<<"time: "<<time<<endl;
            int size = q.size();
            while(size > 0){
                int u = q.front();
                //cout<<endl<<u;
                q.pop();
                visited.insert(u);
                for(int v : adjList[u]){
                    if(visited.find(v) == visited.end()){
                        q.push(v);
                    }
                }
                size--;
            }
            time++;
        }
        return time;
    }
};

/*

adj list

1 - 5, 3
2 - 4
3 - 10, 6, 1
4 - 2, 5, 9
5 - 1, 4
6 - 3
9 - 4
10 - 3

BFS - 3 10 6 1 5 4 2 9
time = 1 + 1 + 1 + 1
*/
191900