LeetCode 3249: Count the Number of Good Nodes Solution

Master LeetCode problem 3249 (Count the Number of Good Nodes), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3249. Count the Number of Good Nodes

Medium

Tree Depth-First Search

Problem Explanation

Explanation

To solve this problem, we can perform a depth-first search (DFS) traversal of the tree starting from the root node. During the traversal, we can keep track of the size of each subtree rooted at a particular node. If all the sizes of the subtrees rooted at the children of a node are the same, then that node is considered a good node.

We can implement the DFS traversal using recursion. At each node, we calculate the sizes of the subtrees rooted at its children and check if they are all the same. If they are, we increment the count of good nodes.

Time Complexity:

The time complexity of this approach is O(n), where n is the number of nodes in the tree.

Space Complexity:

The space complexity of this approach is O(n) for the recursive call stack.

Solution Code

class Solution {
    int count = 0;
    
    public int countGoodNodes(int n, int[][] edges) {
        List<List<Integer>> adjList = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adjList.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            adjList.get(edge[0]).add(edge[1]);
            adjList.get(edge[1]).add(edge[0]);
        }
        
        dfs(0, -1, adjList, new int[n]);
        
        return count;
    }
    
    private int[] dfs(int node, int parent, List<List<Integer>> adjList, int[] subtreeSizes) {
        int currentSize = 1;
        for (int child : adjList.get(node)) {
            if (child != parent) {
                int[] childSizes = dfs(child, node, adjList, subtreeSizes);
                currentSize += childSizes[0];
                subtreeSizes[child] = childSizes[0];
            }
        }
        
        boolean goodNode = true;
        for (int child : adjList.get(node)) {
            if (child != parent && subtreeSizes[child] != currentSize - subtreeSizes[child]) {
                goodNode = false;
                break;
            }
        }
        
        if (goodNode) {
            count++;
        }
        
        return new int[] {currentSize, 0};
    }
}

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