<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 ans = 0;
    void graph(TreeNode* root , map<TreeNode * , vector<TreeNode*>> &mp)
    {
        if(!root) return ;
        else
        {
            if(root->left)
            {
                mp[root].push_back(root->left);
                mp[root->left].push_back(root);
                graph(root->left,mp);
            }

            if(root->right)
            {
                mp[root].push_back(root->right);
                mp[root->right].push_back(root);
                graph(root->right,mp);
            }
        }
    }

    void bfs(TreeNode * start , map<TreeNode * , vector<TreeNode*>> &mp , int time)
    {
        ans = max(ans,time);
        int og = start->val;
        start->val = 0;
        for(auto it : mp[start])
        {
            if(it->val>0)
            {
                time++;
                bfs(it,mp,time);
                time--;
            }
        }

        start->val = og;
    }
    
    TreeNode * find_node(int start , TreeNode * root)
    {
        if(!root) return NULL;
        if(start == root->val) return root;
        TreeNode * ans1 = find_node(start,root->left);
        TreeNode * ans2 = find_node(start,root->right);
        if(ans1) return ans1;
        return ans2;
    }

    int amountOfTime(TreeNode* root, int start) {
        map<TreeNode * , vector<TreeNode*>> mp;
        graph(root,mp);
        TreeNode * strt = find_node(start,root);
        int time = 0;
        bfs(strt,mp,time);
        return ans;
    }
};
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