<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(TreeNode* root,int start){
        if(root==nullptr)return root;
        // if(root->left==nullptr and root->right==nullptr and root->val!=start)return nullptr;
        if(root->val==start){
            return root;
        }
        TreeNode* le=nullptr;
        TreeNode* ri=nullptr;
        if(root->left){
            le= get(root->left,start);
        }
        if(root->right){
            ri= get(root->right,start);
        }
        if(le)return le;
        return ri;
    }
    int time(unordered_map<TreeNode*,TreeNode*> &map,TreeNode* st,set<int> &visit){
        int ans=0;
        visit.insert(st->val);
        if(map[st] and visit.find(map[st]->val)==visit.end()){
            ans=max(ans,1+time(map,map[st],visit));
        }
        if(st->left and visit.find(st->left->val)==visit.end()){
            ans=max(ans,1+time(map,st->left,visit));
        }
        if(st->right and visit.find(st->right->val)==visit.end()){
            ans=max(ans,1+time(map,st->right,visit));
        }
        return ans;
    }
    int amountOfTime(TreeNode* root, int start) {
        TreeNode* node=get(root,start);
        unordered_map<TreeNode*,TreeNode*> map;
        queue<TreeNode*> q;
        q.push(root);
        map[root]=nullptr;
        while(!q.empty()){
            auto t=q.front();
            q.pop();
            if(t->left){
                q.push(t->left);
                map[t->left]=t;
            }
            if(t->right){
                q.push(t->right);
                map[t->right]=t;
            }
        }
        set<int> visit;
        return time(map,node,visit);

    }
};
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