<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>

def dfs(i, parent, parent_to_children, parent_to_max2_children, children_to_parent, node_to_max_down_dist, graph):
    children_to_parent[i] = parent 
    parent_to_children[i] = set()
    parent_to_max2_children[i] = []

    cur_max = 0
    for inext in graph.get(i, set()):
        if inext in children_to_parent:
            continue
        parent_to_children[i].add(inext)
        dfs(inext, i, parent_to_children, parent_to_max2_children, children_to_parent, node_to_max_down_dist, graph)
        cur_dist = node_to_max_down_dist[inext] + (1 if inext % 2 == 1 else 2)
        cur_max = max(cur_max, cur_dist)
        parent_to_max2_children[i].append((cur_dist, inext))
        parent_to_max2_children[i].sort(key = lambda x : -x[0])
        if len(parent_to_max2_children[i]) > 2:
            parent_to_max2_children[i].pop()
    node_to_max_down_dist[i] = cur_max

def dfs2(i, parent, parent_to_max2_children, parent_to_children, node_to_max_up_dist):
    max_up_dist = 0
    if parent is not None:
        max_up_dist = node_to_max_up_dist[parent] + (1 if parent % 2 == 1 else 2)
        for dist, c in parent_to_max2_children[parent]:
            if c != i:
                max_up_dist = max(max_up_dist, (1 if parent % 2 == 1 else 2) + dist)
                break

    node_to_max_up_dist[i] = max_up_dist

    for c in  parent_to_children[i]:
        dfs2(c, i, parent_to_max2_children, parent_to_children, node_to_max_up_dist)

    
class Solution:
    def timeTaken(self, edges: List[List[int]]) -> List[int]:
        parent_to_children = {}
        parent_to_max2_children = {}
        children_to_parent = {}

        node_to_max_down_dist = {}
        node_to_max_up_dist = {}
        

        graph = {}
        for s, e in edges:
            if s not in graph:
                graph[s] = set()
            if e not in graph:
                graph[e] = set()

            graph[s].add(e)
            graph[e].add(s)

        dfs(0, None, parent_to_children, parent_to_max2_children, children_to_parent, node_to_max_down_dist, graph)

        dfs2(0, None, parent_to_max2_children, parent_to_children, node_to_max_up_dist)

        nn = len(edges) + 1
        res = [0] * nn
        for i in range(nn):
            res[i] = max(node_to_max_up_dist[i], node_to_max_down_dist[i])
        
        return res


        
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