<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: 
//     void makegraph(unordered_map<int,vector<int>> &adj, int parent , TreeNode* curr){
//         if(curr == NULL){
//             return;
//         }

//         if(parent != -1){
//             adj[curr->val].push_back(parent);
//         }

//         if(curr->left){
//             adj[curr->val].push_back(curr->left->val);
//         }

//         if(curr->right){
//             adj[curr->val].push_back(curr->right->val);
//         }

//         makegraph(adj, curr->val, curr->left);
//         makegraph(adj, curr->val, curr->right);
//     }

//     int amountOfTime(TreeNode* root, int start) {
//         unordered_map<int,vector<int>> adj;
//         makegraph(adj,-1,root);

//         unordered_set<int> visited;
//         queue<int> q;
//         q.push(start);
//         visited.insert(start);

//         int minutes = 0;
//         while(!q.empty()){
//             int n = q.size();

//             while(n--){
//                 int node = q.front();
//                 q.pop();

//                 for(auto adjnode: adj[node]){
//                     if(visited.find(adjnode) == visited.end()){
//                         q.push(adjnode);
//                         visited.insert(adjnode);
//                     }
//                 }
//             }

//             minutes++;
//         }

//         return minutes -1;
//     }
// };

#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <queue>

using namespace std;

/**
 * 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:
    void makegraph(unordered_map<int, vector<int>> &adj, int parent, TreeNode* curr) {
        if (!curr) return;

        if (parent != -1) {
            adj[curr->val].push_back(parent);
            adj[parent].push_back(curr->val); // Ensure bidirectional connection
        }

        if (curr->left) {
            makegraph(adj, curr->val, curr->left);
        }

        if (curr->right) {
            makegraph(adj, curr->val, curr->right);
        }
    }

    int amountOfTime(TreeNode* root, int start) {
        unordered_map<int, vector<int>> adj;
        makegraph(adj, -1, root);

        unordered_set<int> visited;
        queue<int> q;
        q.push(start);
        visited.insert(start);

        int minutes = 0;

        while (!q.empty()) {
            int levelSize = q.size();
            for (int i = 0; i < levelSize; ++i) {
                int node = q.front();
                q.pop();

                for (int neighbor : adj[node]) {
                    if (visited.find(neighbor) == visited.end()) {
                        visited.insert(neighbor);
                        q.push(neighbor);
                    }
                }
            }
            ++minutes;
        }

        return minutes - 1; // Subtract one to account for the extra increment after the last level
    }
};

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