<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
        from collections import deque, defaultdict
        
        adj = [[] for x in range(n)]
        for u, v in edges:
            adj[u].append(v)
            adj[v].append(u)

        times = dict()
        process = []
        visited = [False] * n
        two_arr = [[(x, x, 0), (x, x, 0)] for x in range(n)]
        def dfs(prev, curr):
            visited[curr] = True
            curr_max = 0
            for node in adj[curr]:
                if not visited[node]:
                    temp = dfs(curr, node)
                    curr_max = max(curr_max, temp)

                    
                    times[(curr, node)] = temp
                    if two_arr[curr][0][2] < temp:
                        something = two_arr[curr][0]
                        two_arr[curr][0] = (curr, node, temp)
                        if(something[2] > two_arr[curr][1][2]):
                            two_arr[curr][1] = something
                    elif two_arr[curr][1][2] < temp:
                        two_arr[curr][1] = (curr, node, temp)

            process.append([curr, prev])

            if(curr % 2):
                return curr_max + 1
            else:
                return curr_max + 2

        dfs(-1, 0)
        ans = [0] * n
        m = len(process) - 1
        for i in range(m - 1, -1, -1):
            src = process[i][0]
            dest = process[i][1]

            curr_max = 0
            for temp in two_arr[dest]:
                if(temp[1] != src):
                    curr_max = max(curr_max, temp[2])

            if(dest%2):
                times[src, dest] = curr_max + 1
                temp = curr_max + 1
                if two_arr[src][0][2] < temp:
                    two_arr[src][0] = (src, dest, temp)
                elif two_arr[src][1][2] < temp:
                    two_arr[src][1] = (src, dest, temp)
            else:
                times[src, dest] = curr_max + 2
                temp = curr_max + 2
                if two_arr[src][0][2] < temp:
                    two_arr[src][0] = (src, dest, temp)
                elif two_arr[src][1][2] < temp:
                    two_arr[src][1] = (src, dest, temp)

        #print(process)
        #print(times)
        
        for i in range(n):
            ans[i] = two_arr[i][0][2]
        return(ans)
128271