<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]:
        # https://leetcode.com/problems/time-taken-to-mark-all-nodes/
        # very dificult implementation. I saw the hints before solving it. Inspired the in the exercise of finding the longest path in a tree
        # calculate the solution from the bottom (dp), and the traverse againg top down genereating the solutions from the top
        
        n = 0
        from collections import defaultdict
        adj = defaultdict(list)
        for u, v in edges:
            if u < v: adj[u].append(v)
            else: adj[v].append(u)
            n = max(n, u+1, v+1)
     
        solution = [0] * n
        solutionFromButtomDict = {}
        solutionFromButtom = [0] * n
        solutionFromTop = [0] * n

        def getSolutionFromButtom(node):

            if not node in adj:
                solutionFromButtom[node] = 0
                return 0

            for neighbour in adj[node]:
                solButtom = getSolutionFromButtom(neighbour)
                solButtom = solButtom + 2 if neighbour % 2 == 0 else solButtom + 1
                solutionFromButtomDict[(node, neighbour)] = solButtom
                solutionFromButtom[node] = max(solutionFromButtom[node], solButtom)
            
            return solutionFromButtom[node]


        import heapq
        
        def getSolutionFromTop(node, solTop):
            
            if not node in adj:
                solutionFromTop[node] = solTop
                solution[node] = solutionFromTop[node]
                return

            # get the max and the second max from neighbours
            solNeighbours = []
            for neighbour in adj[node]:
                solNeighbour = solutionFromButtomDict[(node, neighbour)] 
                solNeighbours.append(-solNeighbour)
            heapq.heapify(solNeighbours)
            maxSolutionFromButtom = 0
            maxSolutionFromButtom2 = 0
            if solNeighbours: maxSolutionFromButtom = - heapq.heappop(solNeighbours)
            if solNeighbours: maxSolutionFromButtom2 = - heapq.heappop(solNeighbours)

            for neighbour in adj[node]:
                solNode = solTop
                solNeighbour = solutionFromButtomDict[(node, neighbour)]
                if solNeighbour != maxSolutionFromButtom: solNode = max(solNode, maxSolutionFromButtom)
                else: solNode = max(solNode, maxSolutionFromButtom2)
                solutionFromTop[node] = solNode
                solution[node] = max(solutionFromTop[node], solutionFromButtom[node])

                getSolutionFromTop(neighbour, solNode + 2 if node % 2 == 0 else solNode + 1)


        getSolutionFromButtom(0)
        getSolutionFromTop(0, 0)

        return solution
116290