<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 {
TreeNode* startNode = NULL;
void findstart(TreeNode* root, int start)
{
    if (root == NULL)
        return ;
    
    if (root -> val == start)
    {
        startNode = root;
        return;
    }

    findstart(root->left, start);
    findstart(root->right, start);

}

public:
    int amountOfTime(TreeNode* root, int start) {

        unordered_map<TreeNode*,TreeNode*> par;
        vector<bool> vis(1e5+1, false);
        int ans = 0;

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

        while (!qu.empty())
        {
            TreeNode *ptr = qu.front();
            qu.pop();

            if (ptr->left)
            {
                par[ptr->left] = ptr;
                qu.push(ptr->left);
            }

            if (ptr->right)
            {
                par[ptr->right] = ptr;
                qu.push(ptr->right);
            }
        }

        findstart(root, start);

        queue<pair<TreeNode*, int>> quu;
        quu.push({startNode, 0});
        vis[startNode->val] = true;

        while (!quu.empty())
        {
            TreeNode* node = quu.front().first;
            int lvl = quu.front().second;
            ans = max(ans, lvl);
            quu.pop();

            if ( node->left && !vis[node->left->val])
            {
                vis[node->left->val] = true;
                quu.push({node->left, lvl + 1});
            }

            if ( node->right && !vis[node->right->val])
            {
                vis[node->right->val] = true;
                quu.push({node->right, lvl + 1});
            }

            if ( par.find(node) != par.end() && !vis[par[node]->val])
            {
                vis[par[node]->val] = true;
                quu.push({par[node], lvl + 1});
            }
        }

        return ans;
    }
};
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