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Problem solution · Java

Number Of Ways To Reconstruct A Tree

Number Of Ways To Reconstruct 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
68 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 Number Of Ways To Reconstruct 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

  • 68 lines of Java from the credited upstream file 1719.java.
  • The implementation visibly relies on sequence storage, hash lookup, ordered lookup.
  • 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 codeNumber Of Ways To Reconstruct A Tree · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  public int checkWays(int[][] pairs) {    final int MAX = 501;    Map<Integer, List<Integer>> graph = new HashMap<>();    int[] degrees = new int[MAX];    boolean[][] connected = new boolean[MAX][MAX];     for (int[] pair : pairs) {      final int u = pair[0];      final int v = pair[1];      graph.putIfAbsent(u, new ArrayList<>());      graph.putIfAbsent(v, new ArrayList<>());      graph.get(u).add(v);      graph.get(v).add(u);      ++degrees[u];      ++degrees[v];      connected[u][v] = true;      connected[v][u] = true;    }     // For each node, sort its children by degrees in descending order.    for (final int u : graph.keySet())      graph.get(u).sort(Comparator.comparingInt(a -> - degrees[a]));     final int root = getRoot(degrees, graph.keySet().size());    if (root == -1)      return 0;    if (!dfs(graph, root, degrees, connected, new ArrayList<>(), new boolean[MAX]))      return 0;    return hasMoreThanOneWay ? 2 : 1;  }   private boolean hasMoreThanOneWay = false;   // Returns the root by finding the node with a degree that equals to n - 1.  private int getRoot(int[] degrees, int n) {    for (int i = 1; i < degrees.length; ++i)      if (degrees[i] == n - 1)        return i;    return -1;  }   // Returns true if each node rooted at u is connected to all of its ancestors.  private boolean dfs(Map<Integer, List<Integer>> graph, int u, int[] degrees,                      boolean[][] connected, List<Integer> ancestors, boolean[] seen) {    seen[u] = true;     for (final int ancestor : ancestors)      if (!connected[u][ancestor])        return false;     ancestors.add(u);     for (final int v : graph.get(u)) {      if (seen[v])        continue;      // We can swap u with v, so there are more than one way.      if (degrees[v] == degrees[u])        hasMoreThanOneWay = true;      if (!dfs(graph, v, degrees, connected, ancestors, seen))        return false;    }     ancestors.remove(ancestors.size() - 1);    return true;  }}

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