<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>

from queue import Queue

class Solution:
    def frogPosition(self, n: int, edges: List[List[int]], t: int, target: int) -> float:
        if edges == []:
            if target == 1:return 1
            return 0

        d = {}
        for i in edges:
            d[i[0]] = d.get(i[0] , []) + [i[1]]
            d[i[1]] = d.get(i[1] , []) + [i[0]]
        
        visit = [0]*(n+1)

        q = Queue()     
        q.put([1 , 1])

        for dur in range(t):
            
            l = q.qsize()
            for i in range(l):
                temp = q.get()

                # Count no.of non-visited nodes
                count = 0
                for ele in d[temp[0]]:
                    if visit[ele] == 0: count += 1
            
                if temp[0] == target and count == 0:     
                    # If the target is reached and if its the end node
                    return temp[1]
                
                if visit[temp[0]] != 0:    # if already visited
                    continue
                
                visit[temp[0]] = 1

                for ele in d[temp[0]]:
                    if visit[ele] == 0: q.put([ele , temp[1]*(1/count)])
                
        
        l = q.qsize()
        for i in range(l):
            temp = q.get()
            if temp[0] == target:
                return temp[1]

        return 0
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