<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>

class Solution:
    # by py-is-best-lang
    def countGoodNodes(self, edges: List[List[int]]) -> int:
        n = len(edges) + 1
        adj = [set() for _ in range(n)]
        for u, v in edges:
            adj[u].add(v)
            adj[v].add(u)
        
        # This while loop is a weird bfs-type traversal to build the array p that
        # contains all nodes (0 to n-1) and in which every node's children 
        # always come (not necessarily immediately) after it.
        # E.g. edges = [[5,0],[4,5],[3,4],[1,4],[2,5],[6,5],[6,7]]
        #                0
        #                |
        #                5
        #            /   |   \
        #           6    4    2
        #          /    / \
        #         7    3   1
        # p = [0, 5, 6, 4, 2, 7, 3, 1]
        # Note that the children of node 4, 3 and 1, come after it, but not immediately
        # after it here.
        # It is sort of like the layers follow each other in the p array, but where one
        # layer ends and the following one begins is not specified.
        p = [0]
        i = 0
        while i < len(p):
            node = p[i]
            for child in adj[node]:
                adj[child].remove(node)
                p.append(child)
            i += 1
        
        size = [1] * n
        ans = 0
        # Having the nodes arranged layer-by-layer in the p array ensures that
        # the children are dealt with before their parents.
        for node in p[::-1]:
            child_size = -1
            is_good = True
            for child in adj[node]:
                size[node] += size[child]
                if child_size == -1:
                    child_size = size[child]
                elif child_size != size[child]:
                    is_good = False
            ans += is_good
        return ans
70131