<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:
   TreeNode *get_and_mapping(TreeNode *root, int target, unordered_map<TreeNode*,TreeNode*> &parent)
{
    // its only map the parent relation from target to root TreeNode path
    if (root == NULL)
        return NULL;
    if (root->val == target)
        return root;
    TreeNode *left = get_and_mapping(root->left, target, parent);
    if (left)
    {
        parent[root->left] = root;
        return left;
    }
    TreeNode *right = get_and_mapping(root->right, target, parent);
    if (right)
    {
        parent[root->right] = root;
        return right;
    }
    // both left and right is NULL
    return NULL;
}
void get_time(TreeNode *root, unordered_map<TreeNode*,TreeNode*> &parent, int c, int &ans)
{
    if (root == NULL)
        return;
    if (ans < c)
        ans = c;
    if(root->left && parent.find(root->left)==parent.end())
    get_time(root->left, parent, c + 1, ans);
    if(root->right && parent.find(root->right)==parent.end())
    get_time(root->right, parent, c + 1, ans);
    if(parent.find(root)!=parent.end()) // no mapping of current TreeNode with parent
    //or
    //if(parent[root])
    get_time(parent[root], parent, c + 1, ans);
}
int time_for_burning(TreeNode *root, int target_data)
{
    unordered_map<TreeNode*,TreeNode*> parent;
    // no need to make parent[root]=NULL bz in recusion only it checks it is present or not
    TreeNode *target = get_and_mapping(root, target_data, parent);
    int ans = 0;

    get_time(target, parent, 0, ans);
    return ans;
}

    int amountOfTime(TreeNode* root, int start) {
      return time_for_burning(root,start);
    }
};
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