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Problem solution · Java

Find the Last Marked Nodes in Tree

Find the Last Marked Nodes in 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
85 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 Find the Last Marked Nodes in 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

  • 85 lines of Java from the credited upstream file 3313.java.
  • The implementation visibly relies on sequence storage, cached states.
  • 4 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 codeFind the Last Marked Nodes in Tree · JavaJava
Use this to learn the idea, then write your own version.
class Solution {  // Similar to 3241. Time Taken to Mark All Nodes  public int[] lastMarkedNodes(int[][] edges) {    final int n = edges.length + 1;    int[] ans = new int[n];    List<Integer>[] tree = new List[n];    // dp[i] := the last marked two nodes for subtree rooted at node i, where    // each node contains the time it got marked    Last2[] dp = new Last2[n];     for (int i = 0; i < n; ++i) {      tree[i] = new ArrayList<>();      dp[i] = new Last2();    }     for (int[] edge : edges) {      final int u = edge[0];      final int v = edge[1];      tree[u].add(v);      tree[v].add(u);    }     dfs(tree, 0, /*prev=*/-1, dp);    reroot(tree, 0, /*prev=*/-1, /*last=*/new Node(), dp, ans);    return ans;  }   private record Node(int node, int time) {    Node() {      this(0, 0);    }  }   private record Last2(Node last1, Node last2) {    Last2() {      this(new Node(), new Node());    }  }   // Performs a DFS traversal of the subtree rooted at node `u`, computes the  // time taken to mark all nodes in the subtree, records the last two marked  // nodes, and returns the last marked node.  //  // These values are used later in the rerooting process.  private Node dfs(List<Integer>[] tree, int u, int prev, Last2[] dp) {    Node last1 = new Node(u, 0);    Node last2 = new Node();    for (final int v : tree[u]) {      if (v == prev)        continue;      Node child = dfs(tree, v, u, dp);      final int time = child.time() + 1;      if (time > last1.time) {        last2 = last1;        last1 = new Node(child.node(), time);      } else if (time > last2.time) {        last2 = new Node(child.node(), time);      }    }    dp[u] = new Last2(last1, last2);    return last1;  }   // Reroots the tree at node `u` and updates the answer array, where `last`  // is the last marked node that doesn't go through `u`'s subtree.  private void reroot(List<Integer>[] tree, int u, int prev, Node last, Last2[] dp, int[] ans) {    ans[u] = last.time() > dp[u].last1().time() ? last.node() : dp[u].last1().node();    for (final int v : tree[u]) {      if (v == prev)        continue;      Node newLast = new Node(last.node(), last.time() + 1); // 1 := u -> v      if (dp[u].last1().node() == dp[v].last1().node()) {        final int alternativeTime = 1 + dp[u].last2().time();        if (alternativeTime > newLast.time())          newLast = new Node(dp[u].last2().node(), alternativeTime);      } else {        final int alternativeTime = 1 + dp[u].last1().time();        if (alternativeTime > newLast.time())          newLast = new Node(dp[u].last1().node(), alternativeTime);      }      reroot(tree, v, u, newLast, dp, ans);    }  }}

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