<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 {
private:
    unordered_map<TreeNode*, TreeNode*> par;

    TreeNode* srcNode = nullptr;
    int srcVal = -1;
    void traverse(TreeNode* root, TreeNode* parent) {
        if(!root) return;


        if(root->val == srcVal) srcNode = root;

        par[root] = parent;
        traverse(root->left, root);
        traverse(root->right, root);
    } 

    int bfs() {
        
        int time = 0;
        unordered_map<TreeNode*, bool> vis;

        queue<TreeNode*> q;
        q.push(srcNode);
        vis[srcNode] = 1;

        while(!q.empty()) {
            
            int qSize = q.size();
            while(qSize--) {
                TreeNode* cur = q.front();
                q.pop();

                

                for(TreeNode* adjNode : {cur->left, cur->right, par[cur]}) {
                    if(adjNode == nullptr) continue;
                    if(adjNode && !vis[adjNode]) {
                        vis[adjNode] = 1;
                        q.push(adjNode);
                    }
                }
            }
            time++;
        }
        
        return time-1;
    }
    
public:
    int amountOfTime(TreeNode* root, int start) {
        srcVal = start;
        traverse(root, nullptr);
        return bfs();
    }
};
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