<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]:
        # build graph
        graph = {}
        for edge in edges:
            if edge[0] not in graph:
                graph[edge[0]] = []
            if edge[1] not in graph:
                graph[edge[1]] = []
            graph[edge[0]].append(edge[1])
            graph[edge[1]].append(edge[0])
        
        root = 0
        
        # calulating time needed to traverse the remaining graph for each node
        subRouteTimes = [0 for i in range(len(edges)+1)]
        visited = [False for i in range(len(edges)+1)]
        def dfs(node):
            maxTimes = 0
            visited[node] = True
            for num in graph[node]:
                if visited[num] is False:
                    time = dfs(num)
                    if num % 2 == 1:
                        time += 1
                    else:
                        time += 2
                    if time > maxTimes:
                        maxTimes = time
            subRouteTimes[node] = maxTimes
            return maxTimes
        
        dfs(root)

        visited = [False for i in range(len(edges)+1)]
        times = [0 for i in range(len(edges)+1)]
        # calculate times start from each node
        def dfs2(node):
            # calculate the minimum time needed to traverse the graph
            visited[node] = True
            maxTimes = 0
            for num in graph[node]:
                time = subRouteTimes[num]
                if num % 2 == 0:
                    time += 2
                else:
                    time += 1
                maxTimes = max(maxTimes, time)
            times[node] = maxTimes

            # to get the updated subRouteTimes, we only need to update node
            # since the graph now denote time start from other nodes other than node itself
            # get the first and second longest time that points to node to increase the efficiency
            fMax, sMax = 0, 0
            for num in graph[node]:
                time = subRouteTimes[num] + 1 if num % 2 == 1 else subRouteTimes[num] + 2
                if time > fMax:
                    sMax = fMax
                    fMax = time
                elif time > sMax:
                    sMax = time

            for num in graph[node]:
                if visited[num]:
                    continue

                # if the new root itself is the one with the longest time, we used the second longest instead
                selfTime = subRouteTimes[num] + 1 if num % 2 == 1 else subRouteTimes[num] + 2
                otherMax = fMax if selfTime < fMax else sMax
                
                
                prev = subRouteTimes[node]
                # update the graph and do the calculation recursively
                subRouteTimes[node] = otherMax
                dfs2(num)
                # recover the information after traversing the node
                subRouteTimes[node] = prev
        
        dfs2(root)



        return times
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