<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]:
        # construct tree graph, u --> v has weight 1 if v odd else 2
        # find v1 and v2 where v1 --> v2 is the longest possible path in tree 
        # BFS/DFS from v1 and v2 to find distances, take maximum
        
        return solve(edges)
    
def solve(edges):
    AL = get_AL(edges)
    v1 = get_v1(AL)
    v2 = get_v2(AL, v1)
    res = []
    AL = {i:[] for i in range(len(edges) + 1)}
    for u, v in edges: # reverse edges
        w1 = 1 if v % 2 else 2
        w2 = 1 if u % 2 else 2
        AL[u].append((v, w2))
        AL[v].append((u, w1))
    d1 = get_distances(AL, v1)
    d2 = get_distances(AL, v2)
    for i in range(len(d1)):
        res.append(max(d1[i], d2[i]))
    return res
        
def get_AL(edges):
    AL = {i:[] for i in range(len(edges) + 1)}
    for u, v in edges:
        w1 = 1 if v % 2 else 2
        w2 = 1 if u % 2 else 2
        AL[u].append((v, w1))
        AL[v].append((u, w2))
    return AL

def get_v1(AL, start = 0):
    visited = [0 for i in range(len(AL))]
    stack = [(start, 0)]
    v1, longest_path = start, 0
    visited[start] = 1
    while stack:
        u, cost = stack.pop()
        if cost > longest_path:
            v1, longest_path = u, cost
        for v, w in AL[u]:
            if not visited[v]:
                visited[v] = 1 
                stack.append((v, cost + w))
    return v1
        
def get_v2(AL, v1):
    return get_v1(AL, v1)

def get_distances(AL, start):
    visited = [0 for i in range(len(AL))]
    distances = [0 for i in range(len(AL))]
    stack = [(start, 0)]
    visited[start] = 1
    while stack:
        u, cost = stack.pop()
        distances[u] = cost
        for v, w in AL[u]:
            if not visited[v]:
                visited[v] = 1 
                stack.append((v, cost + w))
    return distances
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