<p>There is a tree (i.e., a connected, undirected graph with no cycles) structure country network consisting of <code>n</code> cities numbered from <code>0</code> to <code>n - 1</code> and exactly <code>n - 1</code> roads. The capital city is city <code>0</code>. You are given a 2D integer array <code>roads</code> where <code>roads[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> denotes that there exists a <strong>bidirectional road</strong> connecting cities <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>.</p>

<p>There is a meeting for the representatives of each city. The meeting is in the capital city.</p>

<p>There is a car in each city. You are given an integer <code>seats</code> that indicates the number of seats in each car.</p>

<p>A representative can use the car in their city to travel or change the car and ride with another representative. The cost of traveling between two cities is one liter of fuel.</p>

<p>Return <em>the minimum number of liters of fuel to reach the capital city</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/22/a4c380025e3ff0c379525e96a7d63a3.png" style="width: 303px; height: 332px;" />
<pre>
<strong>Input:</strong> roads = [[0,1],[0,2],[0,3]], seats = 5
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
- Representative<sub>1</sub> goes directly to the capital with 1 liter of fuel.
- Representative<sub>2</sub> goes directly to the capital with 1 liter of fuel.
- Representative<sub>3</sub> goes directly to the capital with 1 liter of fuel.
It costs 3 liters of fuel at minimum. 
It can be proven that 3 is the minimum number of liters of fuel needed.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/11/16/2.png" style="width: 274px; height: 340px;" />
<pre>
<strong>Input:</strong> roads = [[3,1],[3,2],[1,0],[0,4],[0,5],[4,6]], seats = 2
<strong>Output:</strong> 7
<strong>Explanation:</strong> 
- Representative<sub>2</sub> goes directly to city 3 with 1 liter of fuel.
- Representative<sub>2</sub> and representative<sub>3</sub> go together to city 1 with 1 liter of fuel.
- Representative<sub>2</sub> and representative<sub>3</sub> go together to the capital with 1 liter of fuel.
- Representative<sub>1</sub> goes directly to the capital with 1 liter of fuel.
- Representative<sub>5</sub> goes directly to the capital with 1 liter of fuel.
- Representative<sub>6</sub> goes directly to city 4 with 1 liter of fuel.
- Representative<sub>4</sub> and representative<sub>6</sub> go together to the capital with 1 liter of fuel.
It costs 7 liters of fuel at minimum. 
It can be proven that 7 is the minimum number of liters of fuel needed.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/09/27/efcf7f7be6830b8763639cfd01b690a.png" style="width: 108px; height: 86px;" />
<pre>
<strong>Input:</strong> roads = [], seats = 1
<strong>Output:</strong> 0
<strong>Explanation:</strong> No representatives need to travel to the capital city.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>roads.length == n - 1</code></li>
	<li><code>roads[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>roads</code> represents a valid tree.</li>
	<li><code>1 &lt;= seats &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def minimumFuelCost(self, roads: List[List[int]], seats: int) -> int:
        # hint: 1. record the size of each subtree
        # hint: 2. If n people meet on the same node, what is the minimum number of cars needed?
        # 1. My Approach given hint: DFS + BFS
        if not roads:
            return 0

        neighbors = collections.defaultdict(list)
        n = 0                   # record the number of cities
        for n1, n2 in roads:
            n = max(n, max(n1, n2))
            neighbors[n1].append(n2)
            neighbors[n2].append(n1)
        n += 1
        sizes = [-1] * n

        def dfs(node, prev) -> int:
            nonlocal neighbors
            nonlocal sizes
            # IMPORTANT: don't add this base case!!! we are not using real node and our first input node will be 0! 
            #if added this case the function ends immediately.
            # if not node:
            #     return 0
            cur_size = 1
            for neighbor in neighbors[node]:        # size of tree = 1 + sum of subtrees' sizes
                if neighbor != prev:
                    cur_size += dfs(neighbor, node)   
            
            sizes[node] = cur_size                  # record size of subtree to list
            return cur_size
            

        dfs(0, -1)

        # BFS: subtree meet together at subtree root to go up by one level 
        res = 0
        queue = collections.deque([[0, -1]])        # store (curNode, prev_node)
        road_len = 0
        while queue:
            length = len(queue)
            for _  in range(length):
                cur, prev = queue.popleft()
                ncars = sizes[cur] // seats
                if sizes[cur] % seats != 0:
                    ncars += 1
                if cur != 0:        # cur is not starting node               
                    res += ncars
                for neighbor in neighbors[cur]:
                    if neighbor != prev:
                        queue.append([neighbor, cur])
            road_len += 1
        
        return res

        
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