<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) {}
 * };
 */


struct Info {
    bool contains;
    int time;
    int depth;
    Info()
    {
        depth = 0;
        contains = false;
        time = -1;
    }
};

class Solution {
    int ans = 0, depthStart = -1;

    int dfs(TreeNode* root, int start, Info& info) {
        if (!root) {
            return 0;
        }

        Info leftInfo, rightInfo;
        int leftH = dfs(root->left, start, leftInfo);
        int rightH = dfs(root->right, start, rightInfo);
        
        if (root->val == start) {

            info.contains = true;
            info.time = 0;
            info.depth = max(leftH, rightH);
            ans = max(ans, info.depth);

            return 1 + max(leftH, rightH);
        }

        if (leftInfo.contains && root->left) {
            if (leftInfo.time != -1){
                info.time = 1 + leftInfo.time;
                info.depth = leftInfo.depth;
            }
            info.contains = info.contains || true;
        } else if (rightInfo.contains && root->right) {
            if (rightInfo.time != -1) {
                info.time = 1 + rightInfo.time;
                info.depth = rightInfo.depth;
            }
            info.contains = info.contains || true;
        }


        // cout<<root->val<<" "<<leftH<<" "<<rightH<<"\n";
        // cout<<info.time<<" "<<info.contains<<"\n";

        if (info.contains) {
            if (root->left && leftInfo.contains) {
                // cout<<"left: "<<leftInfo.depth<<"\n";
                ans = max(ans, max(info.time + rightH,leftInfo.depth));
            } else if (root->right && rightInfo.contains) {
                // cout<<"right: "<<rightInfo.depth<<"\n";
                ans = max(ans, max(info.time + leftH, rightInfo.depth));
            }
        }
        // cout<<"ans: "<<ans<<"\n";


        return 1 + max(leftH, rightH);
    }
public:
    int amountOfTime(TreeNode* root, int start) {
        Info info;
        dfs(root, start, info);
        return ans;
    }
};
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