<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]:
        n = len(edges) + 1
        neibs = [[] for i in range(n)]
        for x, y in edges:
            neibs[x].append(y)
            neibs[y].append(x)
        maxT = [[0 for j in range(len(i))] for i in neibs]
        ans = [0 for i in range(n)]
        def dfs(node, par, time):
            ans[0] = max(ans[0], time)

            mx = time
            # traverse the neighbors
            for i in range(len(neibs[node])):
                j = neibs[node][i]
                if j == par:
                    continue
                if j % 2 == 0:
                    val = dfs(j, node, time + 2)
                else:
                    val = dfs(j, node, time + 1)
                maxT[node][i] = val - time
                mx = max(mx, val)
            return mx
        dfs(0, -1, 0)
        
        def dfs2(node, par, max_time):
            if par != -1:
                for i in range(len(neibs[node])):
                    if neibs[node][i] == par:
                        maxT[node][i] = max_time
            
            mx = second_mx = None
            add = 2 if node % 2 == 0 else 1
            neighbors  = neibs[node]
            times = maxT[node]
            # print(neighbors, times)
            for i in range(len(times)):
                # print(i, times[i], mx, second_mx) # find max_time and second_max time among all neighbors. 
                if mx == None:
                    # print(1)
                    mx = i
                elif times[i] >= times[mx]:
                    # print(2)
                    second_mx = mx
                    mx = i
                elif second_mx == None:
                    # print(3)
                    second_mx = i
                elif times[i] > times[second_mx]:
                    # print(4)
                    second_mx = i
            
            ans[node] = times[mx]
            # print("neibs, maxT", neibs, maxT)
            for i in range(len(neighbors)): # This is for maximum time from node other than neighbor that is travelled through dfs
                if neighbors[i] != par:
                    if mx == None:
                        continue
                    elif i != mx:
                        dfs2(neighbors[i], node, times[mx] + add)
                    elif second_mx == None:
                        dfs2(neighbors[i], node, add)
                    elif i != second_mx:
                        # print("problem", node, neighbors[i], mx, second_mx, neighbors, times)
                        dfs2(neighbors[i], node, times[second_mx] + add)
        dfs2(0, -1, 0)
        return ans
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