<p>You are given a <strong>tree</strong> (i.e. a connected, undirected graph that has no cycles) <strong>rooted</strong> at node <code>0</code> consisting of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code>. The tree is represented by a <strong>0-indexed</strong> array <code>parent</code> of size <code>n</code>, where <code>parent[i]</code> is the parent of node <code>i</code>. Since node <code>0</code> is the root, <code>parent[0] == -1</code>.</p>

<p>You are also given a string <code>s</code> of length <code>n</code>, where <code>s[i]</code> is the character assigned to node <code>i</code>.</p>

<p>Return <em>the length of the <strong>longest path</strong> in the tree such that no pair of <strong>adjacent</strong> nodes on the path have the same character assigned to them.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/25/testingdrawio.png" style="width: 201px; height: 241px;" />
<pre>
<strong>Input:</strong> parent = [-1,0,0,1,1,2], s = &quot;abacbe&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -&gt; 1 -&gt; 3. The length of this path is 3, so 3 is returned.
It can be proven that there is no longer path that satisfies the conditions. 
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/25/graph2drawio.png" style="width: 201px; height: 221px;" />
<pre>
<strong>Input:</strong> parent = [-1,0,0,0], s = &quot;aabc&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest path where each two adjacent nodes have different characters is the path: 2 -&gt; 0 -&gt; 3. The length of this path is 3, so 3 is returned.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == parent.length == s.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= parent[i] &lt;= n - 1</code> for all <code>i &gt;= 1</code></li>
	<li><code>parent[0] == -1</code></li>
	<li><code>parent</code> represents a valid tree.</li>
	<li><code>s</code> consists of only lowercase English letters.</li>
</ul>

class Solution {
public:
    int dfs(int node,vector<vector<int>>&g,string&s,vector<int>&vis,int& ans){
	    //Making the visited node true or marking it visited
        vis[node]=1;
		//We are creating a temp vector which would contain the length of 
		//Subtree with different char at subTree Node
		vector<int>temp;
		
		//Now we start Traversing the neighboring nodes of the val node.
        for(int child:g[node]){
            if(!vis[child]){
                int val=dfs(child,g,s,vis,ans);
				//if the char at child not equal to char at node then we will store the length of subtree
                if(s[child]!=s[node]){
                    temp.push_back(val);
                }
            }
        }
        int maximum=0,val=0;
        sort(temp.begin(),temp.end(),greater<int>());
        for(int i=0;i<2&&i<temp.size();i++){
            maximum+=temp[i];
            val=max(val,temp[i]);
        }
		
        ans=max(ans,maximum+1);
		//Returning the maximum length of single branch
        return val+1;
    }
    int longestPath(vector<int>& parent, string s) {
        int n=s.size();
        vector<vector<int>>g(n);
		//First step is to convert the edges into the graph
		//so as given undirected graph so we have made 2d vector for graph
        for(int i=1;i<n;i++){
            g[i].push_back(parent[i]);
            g[parent[i]].push_back(i);
        }
		//Second step is to make the visited vector 
		//inorder to keep the count of the node that has been visited.
		
        vector<int>vis(n);
        int res=0;
        int ans=dfs(0,g,s,vis,res);
        
        return max(res,ans);
    }
};
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