<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:
#         d = defaultdict(list)

#         for x, y in edges:
#             d[x].append(y)
#             d[y].append(x)
#         # print(d)

#         ans = 0
#         def dfs(node, time, prob, fa, v):
#             nonlocal ans

#             if time == 0 and node == target:
#                 ans = prob
#                 return True

#             if time and node == target:
#                 ns = d[node]

#                 for x in ns:
#                     if x not in v:
#                         return False

#                 ans = prob
#                 return True

#             if time == 0:
#                 return False

#             neigbours = d[node]
#             s_neigbours = set(neigbours)

#             if fa in s_neigbours:
#                 s_neigbours.remove(fa)
                
#             s_len = len(s_neigbours)
    
#             for neigbour in neigbours:
#                 if neigbour == fa:
#                     continue
#                 v.add(neigbour)
#                 res = dfs(neigbour, time-1, prob*(1/(s_len)), node, v)
#                 if res: return True

#             return False

#         dfs(1, t, 1, -1, set([1]))
#         return ans



class Solution:
    def frogPosition(self, n: int, edges: List[List[int]], t: int, target: int) -> float:
        d = defaultdict(list)

        for x, y in edges:
            d[x].append(y)
            d[y].append(x)

        d[1].append(0) # so no need to check if it root node and we do not need to -1 to get the denominator

        ans = 0
        def dfs(node, time, prob, fa):
            nonlocal ans

            if time == 0 and node == target:
                ans = prob
                return True

            if time and node == target:
                ns = d[node]

                # there is still child to be visited
                if len(ns) > 1:
                    return False

                ans = prob
                return True

            if time == 0:
                return False

            neigbours = d[node]

            for neigbour in neigbours:
                if neigbour == fa:
                    continue
                res = dfs(neigbour, time-1, prob*(1/(len(neigbours)-1)), node)
                if res: return True

            return False

        dfs(1, t, 1, -1)
        return ans




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