<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:
    def __init__(self):
        self.longest_path = 1

    def dfs(self, current_node, children, s):
        # If the node is the only child, return 1 for the current_node itself.
        if current_node not in children:
            return 1

        # Longest and second longest path starting from current_node (does not count the
        # current_node itself).
        longest_chain, second_longest_chain = 0, 0
        for child in children[current_node]:
            # Get the number of nodes in the longest chain starting from the child,
            # including the child.
            longest_chain_starting_from_child = self.dfs(child, children, s)
            # We won't move to the child if it has the same character as the current_node.
            if s[current_node] == s[child]:
                continue
            # Modify the longest_chain and second_longest_chain if longest_chain_starting_from_child
            # is bigger.
            if longest_chain_starting_from_child > longest_chain:
                second_longest_chain = longest_chain
                longest_chain = longest_chain_starting_from_child
            elif longest_chain_starting_from_child > second_longest_chain:
                second_longest_chain = longest_chain_starting_from_child

        # Add "1" for the node itself.
        self.longest_path = max(self.longest_path, longest_chain + second_longest_chain + 1)
        return longest_chain + 1

    def longestPath(self, parent, s):
        n = len(parent)
        children = {}
        # Start from node 1, since root node 0 does not have a parent.
        for i in range(1, n):
            if parent[i] not in children:
                children[parent[i]] = []
            children[parent[i]].append(i)

        self.dfs(0, children, s)

        return self.longest_path

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