<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._max_length = 0
 
    def longestPath(self, parent: List[int], s: str) -> int:
        n = len(parent)
        # childrenSet[i] is the set of all the children of node i.
        childrenSet = [set() for i in range(n)]
        for i in range(1, n):
            childrenSet[parent[i]].add(i)
        
        self.dfsHelperSetMax(0, childrenSet, s)
        return self._max_length

    def dfsHelperSetMax(self, root, childrenSet, s):
        if not childrenSet[root]:
            if self._max_length == 0:
                self._max_length = 1
            return 1
        branchMax = 1
        topBranchesLen = []
        for child in childrenSet[root]:
            branchLen = self.dfsHelperSetMax(child, childrenSet, s)
            if s[child] != s[root]:
                # update maxBranch
                if branchLen + 1 > branchMax:
                    branchMax = branchLen + 1
                # update topBranchesLen
                if not topBranchesLen:
                    topBranchesLen.append(branchLen)
                elif len(topBranchesLen) == 1:
                    topBranchesLen.append(branchLen)
                    topBranchesLen.sort()
                else:
                    # already two, and they are sorted
                    topBranchesLen.append(branchLen)
                    topBranchesLen.sort()
                    # topBranchesLen.popleft() # 'list' object has no attribute 'popleft' !
                    topBranchesLen.pop(0)
        # update self._max_length
        if branchMax > self._max_length:
            self._max_length = branchMax
        if len(topBranchesLen) != 0:
            if len(topBranchesLen) == 1:
                if 1 + topBranchesLen[0] > self._max_length:
                    self._max_length = 1 + topBranchesLen[0]
            else:
                if 1 + topBranchesLen[0] + topBranchesLen[1] > self._max_length:
                    self._max_length = 1 + topBranchesLen[0] + topBranchesLen[1]
        return branchMax

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