<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:
    def timeTaken(self, edges: List[List[int]]) -> List[int]:
        adj = defaultdict(list)
        n = 0
        for x, y in edges:
            adj[x].append(y)
            adj[y].append(x)
            n = max(n, x)
            n = max(n, y)
        n += 1

        n_t_h = [(0, None)]
        n_n_t_h = []
        last_seen = None
        while n_t_h or f_n_t_h:
            f_n_t_h = []
            for node, prev in n_t_h:
                for n_n in adj[node]:
                    if n_n == prev:
                        continue
                    if n_n % 2 == 0:
                        f_n_t_h.append((n_n, node))
                    else:
                        n_n_t_h.append((n_n, node))
                last_seen = node
            n_t_h = n_n_t_h
            n_n_t_h = f_n_t_h
        
        n_t_h = [(last_seen, None)]
        n_n_t_h = []
        last_seen2 = None
        time2 = 0
        to_reach2 = [0 for _ in range(n)]
        while n_t_h or f_n_t_h:
            f_n_t_h = []
            for node, prev in n_t_h:
                to_reach2[node] = time2
                for n_n in adj[node]:
                    if n_n == prev:
                        continue
                    if n_n % 2 == 0:
                        f_n_t_h.append((n_n, node))
                    else:
                        n_n_t_h.append((n_n, node))
                last_seen2 = node
            n_t_h = n_n_t_h
            n_n_t_h = f_n_t_h
            time2 += 1

        n_t_h = [(last_seen2, None)]
        n_n_t_h = []
        time1 = 0
        to_reach1 = [0 for _ in range(n)]
        while n_t_h or f_n_t_h:
            f_n_t_h = []
            for node, prev in n_t_h:
                to_reach1[node] = time1
                for n_n in adj[node]:
                    if n_n == prev:
                        continue
                    if n_n % 2 == 0:
                        f_n_t_h.append((n_n, node))
                    else:
                        n_n_t_h.append((n_n, node))
            n_t_h = n_n_t_h
            n_n_t_h = f_n_t_h
            time1 += 1


        minus1 = int(last_seen % 2 == 0)
        minus2 = int(last_seen2 % 2 == 0)
        result = []
        for i in range(n):
            minus = int(i % 2 == 0)
            time1 = to_reach1[i] + minus2 - minus
            time2 = to_reach2[i] + minus1 - minus
            result.append(max(time1, time2))
        
        return result
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