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Problem solution · C++

Find the Last Marked Nodes in Tree

Find the Last Marked Nodes in Tree: a C++ 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
82 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

  • 82 lines of C++ from the credited upstream file 3313.cpp.
  • The implementation visibly relies on sequence storage, cached states.
  • 3 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 · C++C++
Use this to learn the idea, then write your own version.
struct Node {  int node = 0;  // the node number  int time = 0;  // the time it got marked}; struct Last2 {  Node last1;  // the last marked node  Node last2;  // the second last marked node}; class Solution { public:  // Similar to 3241. Time Taken to Mark All Nodes  vector<int> lastMarkedNodes(vector<vector<int>>& edges) {    const int n = edges.size() + 1;    vector<int> ans(n);    vector<vector<int>> tree(n);    // dp[i] := the last marked two nodes for subtree rooted at node i, where    // each node contains the time it got marked    vector<Last2> dp(n);     for (const vector<int>& edge : edges) {      const int u = edge[0];      const int v = edge[1];      tree[u].push_back(v);      tree[v].push_back(u);    }     dfs(tree, 0, /*prev=*/-1, dp);    reroot(tree, 0, /*prev=*/-1, /*last=*/Node(), dp, ans);    return ans;  }  private:  // 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.  Node dfs(const vector<vector<int>>& tree, int u, int prev,           vector<Last2>& dp) {    Node last1(u, 0);    Node last2;    for (const int v : tree[u]) {      if (v == prev)        continue;      Node child = dfs(tree, v, u, dp);      const int time = child.time + 1;      if (time > last1.time) {        last2 = last1;        last1 = Node(child.node, time);      } else if (time > last2.time) {        last2 = Node(child.node, time);      }    }    dp[u] = {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.  void reroot(const vector<vector<int>>& tree, int u, int prev,              const Node& last, vector<Last2>& dp, vector<int>& ans) {    ans[u] = last.time > dp[u].last1.time ? last.node : dp[u].last1.node;    for (const int v : tree[u]) {      if (v == prev)        continue;      Node newLast(last.node, last.time + 1);      if (dp[u].last1.node == dp[v].last1.node) {        const int alternativeTime = 1 + dp[u].last2.time;        if (alternativeTime > newLast.time)          newLast = Node(dp[u].last2.node, alternativeTime);      } else {        const int alternativeTime = 1 + dp[u].last1.time;        if (alternativeTime > newLast.time)          newLast = Node(dp[u].last1.node, alternativeTime);      }      reroot(tree, v, u, newLast, dp, ans);    }  }};

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