<p>There exists an undirected and initially unrooted tree with <code>n</code> nodes indexed from <code>0</code> to <code>n - 1</code>. You are given the integer <code>n</code> and a 2D integer array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>

<p>Each node has an associated price. You are given an integer array <code>price</code>, where <code>price[i]</code> is the price of the <code>i<sup>th</sup></code> node.</p>

<p>The <strong>price sum</strong> of a given path is the sum of the prices of all nodes lying on that path.</p>

<p>The tree can be rooted at any node <code>root</code> of your choice. The incurred <strong>cost</strong> after choosing <code>root</code> is the difference between the maximum and minimum <strong>price sum</strong> amongst all paths starting at <code>root</code>.</p>

<p>Return <em>the <strong>maximum</strong> possible <strong>cost</strong></em> <em>amongst all possible root choices</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/12/01/example14.png" style="width: 556px; height: 231px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[0,1],[1,2],[1,3],[3,4],[3,5]], price = [9,8,7,6,10,5]
<strong>Output:</strong> 24
<strong>Explanation:</strong> The diagram above denotes the tree after rooting it at node 2. The first part (colored in red) shows the path with the maximum price sum. The second part (colored in blue) shows the path with the minimum price sum.
- The first path contains nodes [2,1,3,4]: the prices are [7,8,6,10], and the sum of the prices is 31.
- The second path contains the node [2] with the price [7].
The difference between the maximum and minimum price sum is 24. It can be proved that 24 is the maximum cost.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/11/24/p1_example2.png" style="width: 352px; height: 184px;" />
<pre>
<strong>Input:</strong> n = 3, edges = [[0,1],[1,2]], price = [1,1,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The diagram above denotes the tree after rooting it at node 0. The first part (colored in red) shows the path with the maximum price sum. The second part (colored in blue) shows the path with the minimum price sum.
- The first path contains nodes [0,1,2]: the prices are [1,1,1], and the sum of the prices is 3.
- The second path contains node [0] with a price [1].
The difference between the maximum and minimum price sum is 2. It can be proved that 2 is the maximum cost.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>edges</code> represents a valid tree.</li>
	<li><code>price.length == n</code></li>
	<li><code>1 &lt;= price[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def maxOutput(self, n: int, edges: List[List[int]], price: List[int]) -> int:
        # construct a graph
        self.graph = defaultdict(list)
        self.price = price
        for frm, to in edges:
            self.graph[frm].append(to)
            self.graph[to].append(frm)
        # root at 0, calculate dp[index]: the max price sum for subtree rooted at index
        self.dp = dict()
        self.calculate(0, -1)


        # rerooting
        # def reroot(node, parent, parentContribution): 
        # if node is taken as root, update cost
        self.cost = 0
        self.reroot(0, -1, 0)
        return self.cost



    def reroot(self, curr, parent, parentContribution):
        # reroot at each of curr.children
        for nxt in self.graph[curr]:
            if nxt == parent:
                self.cost = max(self.cost, parentContribution)
            else:
                self.cost = max(self.cost, self.dp[nxt])

        # recursion for children
        # find parentContribution
        hp = [] # min heap
        for nxt in self.graph[curr]:
            if nxt == parent:
                continue
            heapq.heappush(hp, (self.dp[nxt], nxt))
            if len(hp)>2:
                heapq.heappop(hp)
     
        
        for nxt in self.graph[curr]:
            if nxt == parent: continue
            if len(hp) == 1:
                 self.reroot(nxt, curr, self.price[curr]+ parentContribution)
            else:
                if hp[1][1] != nxt:
                    self.reroot(nxt, curr, self.price[curr]+ max(parentContribution, hp[1][0]))
                else:
                    self.reroot(nxt, curr, self.price[curr]+ max(parentContribution, hp[0][0]))
                
            
    def calculate(self, curr, parent):
        # return max price
        price = 0
        for nxt in self.graph[curr]:
            if nxt == parent: continue
            price = max(price, self.calculate(nxt, curr))
        self.dp[curr] = price + self.price[curr]
        return self.dp[curr]



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