<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:
    // declared here, no need to pass in recursive functions
    vector<int>* children; 
    string s;
    int maxSoFar = 1;

    int recur(int node)
    {
        int n = children[node].size();

        int first = 0;      // first highest path in my subtree
        int second = 0;     // second hihghest path in my subtree

        for(int i = 0; i<n; i++)
        { 
            int ele = children[node][i];
            int smallAns = recur(ele);
            
            if(s[ele] == s[node])   // no need to consider if child has same aplphabet
                continue;
            
            // compete candidate soln with first and second highest
            if(smallAns < second)
                continue;
            
            // now candt >= second
            if(smallAns < first)
            {
                second = smallAns;
                continue;
            }
            
            // now candt >= first
            second = first;
            first = smallAns;
        }

        int myGlobAns = first + second + 1;
        if(myGlobAns > maxSoFar)
            maxSoFar = myGlobAns;
        
        return first + 1;   // retuned answer is diff from myGlobAns
    }
    int longestPath(vector<int>& parent, string s) {
        int n = parent.size();

        // these are global/ class level, no need to pass in functions
        this->s = s;
        children = new vector<int>[n];
        
        // converting parent array to children array
        for(int i = 1; i<n; i++)
            children[parent[i]].push_back(i);

        // just call the function with starting node
        recur(0);
        return maxSoFar;
    }
};
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