<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:

    void dfs(int node,int parent,vector<int> adj[],string& s,int& ans,vector<int>& dp){
        dp[node]=1;
        int a=-1,b=-1;
        for(auto &i:adj[node]){
            if(i==parent) continue;
            dfs(i,node,adj,s,ans,dp);
            if(dp[i]>=a && s[i]!=s[node]){
                b=a;
                a=dp[i];
            }
            else if(dp[i]>=b && s[i]!=s[node]){
                b=dp[i];
            }
            // dp[node]=max(dp[node],(s[node]==s[i]?0:1)+dp[i]);
            if(s[i]!=s[node]){
                dp[node]=max(dp[node],1+dp[i]);
            }
        }
        if(a!=-1 && b!=-1){
            ans=max(ans,a+b+1);
        }
    }

    int longestPath(vector<int>& parent, string s) {
        int n=parent.size();
        vector<int> adj[n];
        for(int i=1;i<n;i++){
            int u=i;
            int v=parent[i];
            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        vector<int> dp(n);
        int ans=-1;
        dfs(0,-1,adj,s,ans,dp);
        for(int i=0;i<n;i++){
            // cout<<i<<"--> "<<dp[i]<<"\n";
            ans=max(ans,dp[i]);
        }
        return ans;
    }
};
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