<p>There is an <strong>undirected</strong> tree with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>, and rooted at node <code>0</code>. You are given a 2D integer array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>

<p>A node is <strong>good</strong> if all the <span data-keyword="subtree">subtrees</span> rooted at its children have the same size.</p>

<p>Return the number of <strong>good</strong> nodes in the given tree.</p>

<p>A <strong>subtree</strong> of <code>treeName</code> is a tree consisting of a node in <code>treeName</code> and all of its descendants.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]]</span></p>

<p><strong>Output:</strong> <span class="example-io">7</span></p>

<p><strong>Explanation:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/05/26/tree1.png" style="width: 360px; height: 158px;" />
<p>All of the nodes of the given tree are good.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">edges = [[0,1],[1,2],[2,3],[3,4],[0,5],[1,6],[2,7],[3,8]]</span></p>

<p><strong>Output:</strong> <span class="example-io">6</span></p>

<p><strong>Explanation:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/06/03/screenshot-2024-06-03-193552.png" style="width: 360px; height: 303px;" />
<p>There are 6 good nodes in the given tree. They are colored in the image above.</p>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">edges = [[0,1],[1,2],[1,3],[1,4],[0,5],[5,6],[6,7],[7,8],[0,9],[9,10],[9,12],[10,11]]</span></p>

<p><strong>Output:</strong> <span class="example-io">12</span></p>

<p><strong>Explanation:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/08/08/rob.jpg" style="width: 450px; height: 277px;" />
<p>All nodes except node 9 are good.</p>
</div>
</div>

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

<ul>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li>The input is generated such that <code>edges</code> represents a valid tree.</li>
</ul>

from collections import defaultdict
from typing import List

class Solution:
    def countGoodNodes(self, edges: List[List[int]]) -> int:
        # Step 1: Build the adjacency list for the tree
        adj_list = defaultdict(list)
        for src, target in edges:
            adj_list[src].append(target)
            adj_list[target].append(src)  # Since the tree is undirected

        # Step 2: Initialize the counter for good nodes
        self.good_nodes_count = 0

        # Step 3: Define the DFS function
        def dfs(current, parent):
            """
            Returns the size of the subtree rooted at 'current'.
            Increments 'good_nodes_count' if 'current' is a good node.
            """
            # Initialize the size of the current subtree
            subtree_size = 1

            # List to store sizes of all immediate child subtrees
            child_subtree_sizes = []

            # Traverse all adjacent nodes (children)
            for neighbor in adj_list[current]:
                if neighbor == parent:
                    continue  # Skip the parent to prevent revisiting

                # Recursively compute the size of the child subtree
                child_size = dfs(neighbor, current)
                child_subtree_sizes.append(child_size)

                # Add to the current subtree's size
                subtree_size += child_size

            # Check if 'current' is a good node
            if child_subtree_sizes:
                # Use set to check if all child subtree sizes are equal
                if len(set(child_subtree_sizes)) == 1:
                    self.good_nodes_count += 1
            else:
                # Leaf nodes are trivially good
                self.good_nodes_count += 1

            return subtree_size

        # Step 4: Start DFS from the root node (0) with no parent (-1)
        dfs(0, -1)

        return self.good_nodes_count

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