<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:

     // Approach 1 using parent map

    // int amountOfTime(TreeNode* root, int start) 
    // {
    //    if(root==NULL)
    //    {
    //     return 0;
    //    } 
    //    map<TreeNode*,TreeNode*> parent;
    //    queue<TreeNode*>q;
    //    q.push(root);
    //    TreeNode* Node;
    //    while(!q.empty())
    //    {
    //      auto temp = q.front();
    //      q.pop();
    //      if(temp->val==start)
    //      {
    //         Node=temp;
    //      }
    //      if(temp->left!=NULL)
    //      {
    //         parent[temp->left]=temp;
    //         q.push(temp->left);
    //      }
    //      if(temp->right!=NULL)
    //      {
    //         parent[temp->right]=temp;
    //         q.push(temp->right);
    //      }
    //    }
       
    //    queue<TreeNode*>q1;
    //    q1.push(Node);
    //    map<TreeNode*,int> vis;
    //    vis[Node]=1;
    //    int ans=0;
    //    while(!q1.empty())
    //    {
    //      int n = q1.size();
    //      int change=0;
    //      for(int i=0;i<n;i++)
    //      {
    //         auto temp = q1.front();
    //         q1.pop();

    //         if(temp->left && vis[temp->left]==0)
    //         {
    //            q1.push(temp->left);
    //            vis[temp->left]=1;
    //            change++;
    //         }
    //         if(temp->right && vis[temp->right]==0)
    //         {
    //            q1.push(temp->right);
    //            vis[temp->right]=1;
    //            change++;
    //         }
    //         if(parent[temp] && vis[parent[temp]]==0)
    //         {
    //            q1.push(parent[temp]);
    //            vis[parent[temp]]=1;
    //            change++;  
    //         }
    //      }
    //      if(change) ans++;
    //    }
    //    return ans;
    // }

    // Approach 2 by doing something in height

    int maxi = 0;
    int solve(TreeNode* root, int start)
    {
        if(root==NULL) return 0;
        
        int lh = solve(root->left,start);
        int rh = solve(root->right,start);

        if(root->val == start)
        {
            maxi = max(lh,rh);
            return -1;
        }
        else if(lh>=0 && rh>=0)
        {
            return max(lh,rh)+1;
        }
        else 
        {
            int d = abs(lh) + abs(rh);
            maxi = max(maxi,d);
            return min(lh,rh)-1;
        }
        //return 0;
    }

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