<p>There is an undirected tree with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>, rooted at node <code>0</code>. You are given 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>At every node <code>i</code>, there is a gate. You are also given an array of even integers <code>amount</code>, where <code>amount[i]</code> represents:</p>

<ul>
	<li>the price needed to open the gate at node <code>i</code>, if <code>amount[i]</code> is negative, or,</li>
	<li>the cash reward obtained on opening the gate at node <code>i</code>, otherwise.</li>
</ul>

<p>The game goes on as follows:</p>

<ul>
	<li>Initially, Alice is at node <code>0</code> and Bob is at node <code>bob</code>.</li>
	<li>At every second, Alice and Bob <b>each</b> move to an adjacent node. Alice moves towards some <strong>leaf node</strong>, while Bob moves towards node <code>0</code>.</li>
	<li>For <strong>every</strong> node along their path, Alice and Bob either spend money to open the gate at that node, or accept the reward. Note that:
	<ul>
		<li>If the gate is <strong>already open</strong>, no price will be required, nor will there be any cash reward.</li>
		<li>If Alice and Bob reach the node <strong>simultaneously</strong>, they share the price/reward for opening the gate there. In other words, if the price to open the gate is <code>c</code>, then both Alice and Bob pay&nbsp;<code>c / 2</code> each. Similarly, if the reward at the gate is <code>c</code>, both of them receive <code>c / 2</code> each.</li>
	</ul>
	</li>
	<li>If Alice reaches a leaf node, she stops moving. Similarly, if Bob reaches node <code>0</code>, he stops moving. Note that these events are <strong>independent</strong> of each other.</li>
</ul>

<p>Return<em> the <strong>maximum</strong> net income Alice can have if she travels towards the optimal leaf node.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/29/eg1.png" style="width: 275px; height: 275px;" />
<pre>
<strong>Input:</strong> edges = [[0,1],[1,2],[1,3],[3,4]], bob = 3, amount = [-2,4,2,-4,6]
<strong>Output:</strong> 6
<strong>Explanation:</strong> 
The above diagram represents the given tree. The game goes as follows:
- Alice is initially on node 0, Bob on node 3. They open the gates of their respective nodes.
  Alice&#39;s net income is now -2.
- Both Alice and Bob move to node 1. 
&nbsp; Since they reach here simultaneously, they open the gate together and share the reward.
&nbsp; Alice&#39;s net income becomes -2 + (4 / 2) = 0.
- Alice moves on to node 3. Since Bob already opened its gate, Alice&#39;s income remains unchanged.
&nbsp; Bob moves on to node 0, and stops moving.
- Alice moves on to node 4 and opens the gate there. Her net income becomes 0 + 6 = 6.
Now, neither Alice nor Bob can make any further moves, and the game ends.
It is not possible for Alice to get a higher net income.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/10/29/eg2.png" style="width: 250px; height: 78px;" />
<pre>
<strong>Input:</strong> edges = [[0,1]], bob = 1, amount = [-7280,2350]
<strong>Output:</strong> -7280
<strong>Explanation:</strong> 
Alice follows the path 0-&gt;1 whereas Bob follows the path 1-&gt;0.
Thus, Alice opens the gate at node 0 only. Hence, her net income is -7280. 
</pre>

<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;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li><code>edges</code> represents a valid tree.</li>
	<li><code>1 &lt;= bob &lt; n</code></li>
	<li><code>amount.length == n</code></li>
	<li><code>amount[i]</code> is an <strong>even</strong> integer in the range <code>[-10<sup>4</sup>, 10<sup>4</sup>]</code>.</li>
</ul>

class Solution:
    def get_bob_pos(self, n, graph, bob):
        prevs = [-1] * n
        q = deque([(bob, -1)])  # (node, prev)

        while len(q) > 0:
            node, prev = q.popleft()

            if node == 0:
                break
            
            for nxt in graph[node]:
                if nxt == prev:
                    continue
                
                prevs[nxt] = node
                q.append((nxt, node))

        #
        ans = [0]
        node = 0
        while prevs[node] != -1:
            ans.append(prevs[node])
            node = prevs[node]

        return ans
    
    def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
        n = len(edges) + 1
        graph = [[] for _ in range(n)]
        for a, b in edges:
            graph[a].append(b)
            graph[b].append(a)

        bob_pos = self.get_bob_pos(n, graph, bob)[::-1]

        opened = [False] * n
        q = deque([(0, -1, 0)])  # (now_node, pre_node, pay)
        t = 0

        ans = float("-inf")

        while len(q) > 0:
            sz = len(q)
            for _ in range(sz):
                node, pre_node, pay = q.popleft()

                if not opened[node]:
                    if t < len(bob_pos) and bob_pos[t] == node:
                        pay += int(amount[node] / 2)
                    else:
                        pay += amount[node]
                
                opened[node] = True

                if len(graph[node]) == 1 and node != 0:
                    ans = max(ans, pay)
                
                for nxt in graph[node]:
                    if nxt == pre_node:
                        continue
                    
                    q.append((nxt, node, pay))


            if t < len(bob_pos):
                opened[bob_pos[t]] = True

            t += 1

        return ans if ans != float("-inf") else -1
57000