<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 getParent(TreeNode*root, map<TreeNode*, TreeNode*>& ans){
        if (root == NULL){
            return;
        }
        if (root -> left){
            ans[root -> left] = root;
        }
        if (root -> right){
            ans[root -> right] = root;
        }

        getParent(root -> left, ans);
        getParent(root -> right, ans);
    }

    TreeNode* getTargetNode(TreeNode* root, int start){
        if (root == NULL){
            return NULL;
        }

        if (root -> val == start){
            return root;
        }

        TreeNode* left = getTargetNode(root -> left, start);
        TreeNode* right = getTargetNode(root -> right, start);

        if (left == NULL){
            return right;
        }
        return left;
    }

    int solve(TreeNode* target, map<TreeNode*, TreeNode*> parentChild){
        
        queue<TreeNode*> q;
        map<TreeNode*, bool> check;
        int ans = 0;

        q.push(target);
        check[target] = true;

        while (!q.empty()){
            int size = q.size();
            bool flag = false;
            for (int i = 0; i < size; i++){

                TreeNode* front = q.front();
                q.pop();

                if (front -> left && !check[front -> left]){
                    q.push(front -> left);
                    check[front -> left] = true;
                    flag = true;
                }

                if (front -> right && !check[front -> right]){
                    q.push(front -> right);
                    check[front -> right] = true;
                    flag = true;
                }

                if (parentChild[front] && !check[parentChild[front]]){
                    q.push(parentChild[front]);
                    check[parentChild[front]] = true;
                    flag = true;
                }

            }
            if (flag){
                ans++;
            }
        }
        return ans;
    }

    int amountOfTime(TreeNode* root, int start) {
        ios_base::sync_with_stdio(false);
	    cin.tie(NULL);
        
        int time = 0;

        if (root == NULL){
            return time;
        }

        map<TreeNode*, TreeNode*> parentChild;
        getParent(root, parentChild);

        TreeNode* targetNode = getTargetNode(root, start);

        time = solve(targetNode, parentChild);
        return time;
    }
};
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