<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>

static int io_opt = []() {
  std::ios::sync_with_stdio(false);
  return 0;
}();

/**
 * 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) {}
 * };
 */

enum {
  UNKNOWN,
  HAS_START_NODE,
};

class Solution {
 public:
  int amountOfTime(TreeNode* root, int start) {
    ans_ = 0;
    search(root, start);

    return ans_;
  }

 protected:
  // <status, path>
  pair<int, int> search(TreeNode* node, int start) {
    if (nullptr == node) return {UNKNOWN, -1};
    auto ld = search(node->left, start);
    auto rd = search(node->right, start);
    if (start == node->val) {
      ans_ = max(ans_, max(ld.second, rd.second) + 1);

      return {HAS_START_NODE, 0};
    }
    if (ld.first == HAS_START_NODE) {
      ans_ = max(ans_, ld.second + rd.second + 2);
      return {HAS_START_NODE, ld.second + 1};
    } else if (rd.first == HAS_START_NODE) {
      ans_ = max(ans_, ld.second + rd.second + 2);
      return {HAS_START_NODE, rd.second + 1};
    }

    return {UNKNOWN, max(ld.second, rd.second) + 1};
  }

 private:
  int ans_;
};
93900