<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
        neighbour = collections.defaultdict(list)
        for u, v in edges :
            neighbour[u].append(v)
            neighbour[v].append(u)
            
        times = {}
        memory = {}
        father_id = {}
        def solve(fp=-1, np=0) :
            to_ret_base = (1 if np % 2 == 1 else 2)
            to_ret_add = 0
            memt = []
            for ison, sont in enumerate(neighbour[np]) :
                if sont == fp :
                    memt.append(-1)
                    father_id[np]=ison
                    continue
                solve_next = solve(np, sont)
                memt.append(solve_next)
                to_ret_add = max(to_ret_add, solve_next)
            memory[np] = memt
            times[(fp, np)] = to_ret_base+to_ret_add
            return to_ret_base+to_ret_add
        
        
        memory[0] = []
        for ni in neighbour[0] :
            memory[0].append(solve(0, ni))
        
        # 这里不预存也是会超时。
        @functools.lru_cache(None)
        def get_sorted(fp) :
            return sorted([[t, i] for i, t in enumerate(memory[fp])], reverse=True)
    
        def solve2(fp=-1, np=0) :
            # memt = sorted([[t, i] for i, t in enumerate(memory[fp])], reverse=True)
            memt = get_sorted(fp)
            if len(memt) == 1 :
                maxt = 0
            elif not np == neighbour[fp][memt[0][1]] :
                maxt = memt[0][0]
            else :
                maxt = memt[1][0]
                
            times[(np, fp)] = maxt + (1 if fp % 2 == 1 else 2)
            memory[np][father_id[np]] = times[(np, fp)]
            
            for sont in neighbour[np] :
                if not sont == fp :
                    solve2(np, sont)
                
        for ni in neighbour[0] :
            solve2(0, ni)
            
        to_ret = []
        for i in range(n) :
            to_ret.append(max([times[(i, ni)] for ni in neighbour[i]]))
        return to_ret

164215