<p>There exists an <strong>undirected</strong> tree with <code>n</code> nodes numbered <code>0</code> to <code>n - 1</code>. You are given a 2D integer array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> indicates that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> in the tree.</p>

<p>Initially, <strong>all</strong> nodes are <strong>unmarked</strong>. For each node <code>i</code>:</p>

<ul>
	<li>If <code>i</code> is odd, the node will get marked at time <code>x</code> if there is <strong>at least</strong> one node <em>adjacent</em> to it which was marked at time <code>x - 1</code>.</li>
	<li>If <code>i</code> is even, the node will get marked at time <code>x</code> if there is <strong>at least</strong> one node <em>adjacent</em> to it which was marked at time <code>x - 2</code>.</li>
</ul>

<p>Return an array <code>times</code> where <code>times[i]</code> is the time when all nodes get marked in the tree, if you mark node <code>i</code> at time <code>t = 0</code>.</p>

<p><strong>Note</strong> that the answer for each <code>times[i]</code> is <strong>independent</strong>, i.e. when you mark node <code>i</code> all other nodes are <em>unmarked</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">edges = [[0,1],[0,2]]</span></p>

<p><strong>Output:</strong> [2,4,3]</p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/01/screenshot-2024-06-02-122236.png" style="width: 500px; height: 241px;" /></p>

<ul>
	<li>For <code>i = 0</code>:

	<ul>
		<li>Node 1 is marked at <code>t = 1</code>, and Node 2 at <code>t = 2</code>.</li>
	</ul>
	</li>
	<li>For <code>i = 1</code>:
	<ul>
		<li>Node 0 is marked at <code>t = 2</code>, and Node 2 at <code>t = 4</code>.</li>
	</ul>
	</li>
	<li>For <code>i = 2</code>:
	<ul>
		<li>Node 0 is marked at <code>t = 2</code>, and Node 1 at <code>t = 3</code>.</li>
	</ul>
	</li>
</ul>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">edges = [[0,1]]</span></p>

<p><strong>Output:</strong> [1,2]</p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/01/screenshot-2024-06-02-122249.png" style="width: 500px; height: 257px;" /></p>

<ul>
	<li>For <code>i = 0</code>:

	<ul>
		<li>Node 1 is marked at <code>t = 1</code>.</li>
	</ul>
	</li>
	<li>For <code>i = 1</code>:
	<ul>
		<li>Node 0 is marked at <code>t = 2</code>.</li>
	</ul>
	</li>
</ul>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">edges = </span>[[2,4],[0,1],[2,3],[0,2]]</p>

<p><strong>Output:</strong> [4,6,3,5,5]</p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/06/03/screenshot-2024-06-03-210550.png" style="height: 266px; width: 500px;" /></p>
</div>

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

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= edges[i][0], edges[i][1] &lt;= n - 1</code></li>
	<li>The input is generated such that <code>edges</code> represents a valid tree.</li>
</ul>

class Solution:
    # Returns max distance from cur to anywhere EXCEPT via prev
    # does NOT count the distance from prev to cur
    def mWP(self, prev: int, cur: int) -> int:
        # if self.maxWithoutParents[cur] > -1:
        #     return self.maxWithoutParents[cur]
        m = 0
        sm = 0
        for nxt in self.adjList[cur]:
            if nxt == prev:
                continue
            dist = (nxt + 1) % 2 + 1
            dist += self.mWP(cur, nxt)
            if sm < dist:
                sm = dist
            if m < sm:
                m, sm = sm, m
        self.maxWithoutParents[cur] = m
        self.secondMaxWithoutParents[cur] = sm
        return m

    # Determines the max distance from cur to anywhere
    # 
    def mP(self, prev: int, cur: int):
        prevCur = (cur + 1) % 2 + 1
        curPrev = (prev + 1) % 2 + 1
        
        # print(self.maxWithoutParents)
        longestPrevCur = self.maxWithoutParents[cur] + prevCur
        # print(f"{prev=} {cur=} {longestPrevCur=}")
        
        if self.maxWithoutParents[prev] == longestPrevCur:
            longestCurPrev = self.secondMaxWithoutParents[prev] + curPrev
        else:
            longestCurPrev = self.maxWithoutParents[prev] + curPrev
        
        if self.secondMaxWithoutParents[cur] < longestCurPrev:
            self.secondMaxWithoutParents[cur] = longestCurPrev
        if self.maxWithoutParents[cur] < self.secondMaxWithoutParents[cur]:
            self.maxWithoutParents[cur], self.secondMaxWithoutParents[cur] = self.secondMaxWithoutParents[cur], self.maxWithoutParents[cur]
    
        for nxt in self.adjList[cur]:
            if prev == nxt:
                continue
            self.mP(cur, nxt)
            
    
    def timeTaken(self, edges: List[List[int]]) -> List[int]:
        adjList = {}
        for n1, n2 in edges:
            if n1 not in adjList:
                adjList[n1] = []
            adjList[n1].append(n2)
            if n2 not in adjList:
                adjList[n2] = []
            adjList[n2].append(n1)
        self.adjList = adjList
        
        nodes = len(adjList)
        self.maxWithoutParents = [-1 for _ in range(nodes)]
        self.secondMaxWithoutParents = [-1 for _ in range(nodes)]
        
        self.mWP(0, 0)
        print(self.maxWithoutParents, self.secondMaxWithoutParents)
        
        # self.result = [-1 for _ in range(nodes)]
        # self.result[0] = self.maxWithoutParents[0]
        for nxt in self.adjList[0]:
            self.mP(0, nxt)
        
        return self.maxWithoutParents
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