<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:
        # return: prob of frog on target after exactly t jumps
        # tree traversal BFS
        # p(child) = p(parent) / children_count
        # found -> return p(child) if jump == target or child has no chlidren, else 0

        # Create adjacency list for more efficient children retrieval
        E = {i: [] for i in range(1, n+1)}
        for edge in edges:
            E[edge[0]].append(edge[1])
            E[edge[1]].append(edge[0])

        # Keep track of visited nodes
        visited = [False for i in range(n+1)]

        # Stack of nodes to be examined
        stack = [[1, 1, 0]] # node, prob, jump

        while stack:
            crt_node, crt_p, crt_jump = stack.pop()
            # Not found
            if crt_jump > t:
                return 0

            visited[crt_node] = True
            crt_children = [n for n in E[crt_node] if not visited[n]]
            crt_children_count = len(crt_children)
            
            if crt_node == target:
                return crt_p if crt_jump == t or crt_children_count == 0 else 0

            # End of branching
            if crt_children_count == 0:
                continue

            p_children = crt_p / len(crt_children)

            for child in crt_children:
                stack.insert(0, [child, p_children, crt_jump+1])

        return 0
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