<p>Given an undirected tree consisting of <code>n</code> vertices numbered from <code>1</code> to <code>n</code>. A frog starts jumping from <strong>vertex 1</strong>. In one second, the frog jumps from its current vertex to another <strong>unvisited</strong> vertex if they are directly connected. The frog can not jump back to a visited vertex. In case the frog can jump to several vertices, it jumps randomly to one of them with the same probability. Otherwise, when the frog can not jump to any unvisited vertex, it jumps forever on the same vertex.</p>

<p>The edges of the undirected tree are given in the array <code>edges</code>, where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> means that exists an edge connecting the vertices <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>.</p>

<p><em>Return the probability that after <code>t</code> seconds the frog is on the vertex <code>target</code>. </em>Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/21/frog1.jpg" style="width: 338px; height: 304px;" />
<pre>
<strong>Input:</strong> n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 2, target = 4
<strong>Output:</strong> 0.16666666666666666 
<strong>Explanation:</strong> The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 probability to the vertex 2 after <strong>second 1</strong> and then jumping with 1/2 probability to vertex 4 after <strong>second 2</strong>. Thus the probability for the frog is on the vertex 4 after 2 seconds is 1/3 * 1/2 = 1/6 = 0.16666666666666666. 
</pre>

<p><strong class="example">Example 2:</strong></p>
<strong><img alt="" src="https://assets.leetcode.com/uploads/2021/12/21/frog2.jpg" style="width: 304px; height: 304px;" /></strong>

<pre>
<strong>Input:</strong> n = 7, edges = [[1,2],[1,3],[1,7],[2,4],[2,6],[3,5]], t = 1, target = 7
<strong>Output:</strong> 0.3333333333333333
<strong>Explanation: </strong>The figure above shows the given graph. The frog starts at vertex 1, jumping with 1/3 = 0.3333333333333333 probability to the vertex 7 after <strong>second 1</strong>. 
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n</code></li>
	<li><code>1 &lt;= t &lt;= 50</code></li>
	<li><code>1 &lt;= target &lt;= n</code></li>
</ul>

class Solution:
    def frogPosition(self, n: int, edges: List[List[int]], t: int, target: int) -> float:
        # 构建邻接表
        nei = defaultdict(set)
        for a, b in edges:
            nei[a].add(b)
            nei[b].add(a)

        visited = set()
        res = 0.0
        
        def dfs(node, probability, time):
            nonlocal res
            # 如果时间用完了或者没有更多时间，并且已经到达目标节点
            if time > t:
                return
            if node == target:
                # 如果在时间内到达了目标节点
                if time == t or len(nei[node] - visited) == 0:
                    res = probability
                return
            
            visited.add(node)
            neighbors = nei[node] - visited  # 排除访问过的节点

            if neighbors:
                # 遍历每个子节点，分配概率
                for neighbor in neighbors:
                    dfs(neighbor, probability / len(neighbors), time + 1)
            else:
                # 如果是叶节点且时间未用完，判断是否在目标节点
                if node == target:
                    res = probability
            visited.remove(node)
        
        # 从节点 1 开始 DFS
        dfs(1, 1.0, 0)
        return res
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