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Problem solution · Java

Count Valid Paths in a Tree

Count Valid Paths in a Tree: a Java solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
57 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Depth-first search

For Count Valid Paths in a Tree, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 57 lines of Java from the credited upstream file 2867.java.
  • The implementation visibly relies on sequence storage.
  • 5 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount Valid Paths in a Tree · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public long countPaths(int n, int[][] edges) {    final boolean[] isPrime = sieveEratosthenes(n + 1);    List<Integer>[] graph = new List[n + 1];     for (int i = 1; i <= n; ++i)      graph[i] = new ArrayList<>();     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      graph[u].add(v);      graph[v].add(u);    }     dfs(graph, 1, /*prev=*/-1, isPrime);    return ans;  }   private long ans = 0;   private Pair<Long, Long> dfs(List<Integer>[] graph, int u, int prev, boolean[] isPrime) {    long countZeroPrimePath = isPrime[u] ? 0 : 1;    long countOnePrimePath = isPrime[u] ? 1 : 0;     for (final int v : graph[u]) {      if (v == prev)        continue;      Pair<Long, Long> pair = dfs(graph, v, u, isPrime);      final long countZeroPrimeChildPath = pair.getKey();      final long countOnePrimeChildPath = pair.getValue();      ans +=          countZeroPrimePath * countOnePrimeChildPath + countOnePrimePath * countZeroPrimeChildPath;      if (isPrime[u]) {        countOnePrimePath += countZeroPrimeChildPath;      } else {        countZeroPrimePath += countZeroPrimeChildPath;        countOnePrimePath += countOnePrimeChildPath;      }    }     return new Pair<>(countZeroPrimePath, countOnePrimePath);  }   private boolean[] sieveEratosthenes(int n) {    boolean[] isPrime = new boolean[n];    Arrays.fill(isPrime, true);    isPrime[0] = false;    isPrime[1] = false;    for (int i = 2; i * i < n; ++i)      if (isPrime[i])        for (int j = i * i; j < n; j += i)          isPrime[j] = false;    return isPrime;  }}

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