<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:
    // Helper function to find the maximum value in the binary tree
    int findMaxValue(TreeNode* root) {
        if (root == nullptr) return 0;  // base case

        int leftMax = findMaxValue(root->left);
        int rightMax = findMaxValue(root->right);

        return max(root->val, max(leftMax, rightMax));  // return maximum value
    }

    // Function to create adjacency list from binary tree
    void createAdj(vector<vector<int>>& adj, TreeNode* root) {
        if (root == nullptr) return;  // base case: root is null

        int node = root->val;
        if (root->left) {
            int left = root->left->val;
            adj[node].push_back(left);
            adj[left].push_back(node);
            createAdj(adj, root->left);  // recursive call for left subtree
        }
        if (root->right) {
            int right = root->right->val;
            adj[node].push_back(right);
            adj[right].push_back(node);
            createAdj(adj, root->right);  // recursive call for right subtree
        }
    }

    int amountOfTime(TreeNode* root, int start) {
        int ans=0;
        if (root == nullptr) return 0;  // no nodes to infect

        // Step 1: Find the maximum node value to determine the size of the adjacency list
        int maxVal = findMaxValue(root);

        // Step 2: Initialize adjacency list, visited array, and distance array
        vector<vector<int>> adj(maxVal + 1);  // Initialize adjacency list of size maxVal + 1
        vector<int> dist(maxVal + 1, INT_MIN);  // Distance from the start node
        vector<bool> visited(maxVal + 1, false);  // Visited array for BFS

        // Step 3: Create adjacency list from binary tree
        createAdj(adj, root);

        // Step 4: Perform BFS to calculate the maximum time to infect all nodes
        queue<int> q;
        q.push(start);
        visited[start] = true;
        dist[start] = 0;

        while (!q.empty()) {
            int node = q.front();
            q.pop();
            for (auto neighbor : adj[node]) {
                if (!visited[neighbor]) {
                    visited[neighbor] = true;
                    dist[neighbor] = dist[node] + 1;
                  
                    q.push(neighbor);
                }
            }
        }

        // Step 5: Find the maximum distance, which is the time needed to infect all nodes
        return *max_element(dist.begin(), dist.end());
    }
};

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