<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>

import math
from collections import deque, defaultdict
import sys
sys.setrecursionlimit(10**5) 
class Solution:
    def mostProfitablePath(self, edges: List[List[int]], bob: int, amount: List[int]) -> int:
        graph = defaultdict(set)
        for u, v in edges:
            graph[u].add(v)
            graph[v].add(u)
        visited = set()
        self.lvl = 0
        self.sum = 0
        self.max = -math.inf
        def remove_bob_income(src, bob, lvl):
            if src == bob:
                self.lvl = lvl
                amount[bob] = 0
                return True
            visited.add(src)
            for child in graph[src]:
                if child not in visited:
                    ans = remove_bob_income(child, bob, lvl + 1)
                    if ans:
                        if self.lvl & 1 and lvl == (self.lvl + 1)//2:
                            amount[src] //= 2
                        elif lvl > (self.lvl // 2):
                            amount[src] = 0
                        else:
                            self.sum += amount[src]   
                        return True
            return False
        
            visited.remove(src)
        
        def dfs(src, income, visited):
            if len(graph[src]) == 1 and src != 0:
                self.max = max(self.max, income)
                return
            visited.add(src)
            for child in graph[src]:
                if child not in visited:
                    dfs(child, income + amount[child], visited)
            visited.remove(src)
                
        remove_bob_income(0, bob, 1)
        dfs(0, amount[0], set())
        return self.max
72600