<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:
    int amountOfTime(TreeNode* root, int target) {
         int ans=INT_MIN;
           if(root==NULL){
               return 0;
            }
        map<int,vector<int>>m;
        queue<TreeNode*>q;
        q.push(root);
        while(!q.empty()){
            int sz=q.size();
            for(int i=0;i<sz;i++){
                TreeNode*a=q.front();
                q.pop();
                if(a->left!=NULL){
                    q.push(a->left);
                    m[a->val].push_back(a->left->val);
                    m[a->left->val].push_back(a->val);
                }if(a->right!=NULL){
                    q.push(a->right);
                    m[a->val].push_back(a->right->val);
                    m[a->right->val].push_back(a->val);
                }
            }
            
        }
        map<int,int>dis;
        for(auto it:m){
            if(it.first!=target){
                dis[it.first]=INT_MAX;
            }else{
                dis[target]=0;
            }
        }
        set<pair<int,int>>s;  //dis,node
        s.insert({0,target});
        while(!s.empty()){
            auto u=*s.begin();
            s.erase(u);
            int dist=u.first;
            int node=u.second;
            for(auto it:m[node]){
                if(dis[it]>1+dis[node]){
                    dis[it]=1+dis[node];
                    s.insert({dis[it],it});
                }
            }
            
        }
        for(auto it:dis){
            ans=max(ans,it.second);
        }
        if(ans==INT_MIN){
            return 0;
        }
        return ans;
    }
};
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