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Problem solution · Python

Find the Last Marked Nodes in Tree

Find the Last Marked Nodes in Tree: a Python 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
90 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

  • 90 lines of Python from the credited upstream file 3313.py.
  • The implementation visibly relies on sequence storage, cached states.
  • No explicit 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 · PythonPython
Use this to learn the idea, then write your own version.
from dataclasses import dataclass  @dataclassclass Node:  node: int = 0  # the node number  time: int = 0  # the time it got marked  class Last2:  def __init__(self, last1: Node = Node(), last2: Node = Node()):    self.last1 = last1  # the last marked node    self.last2 = last2  # the second last marked node  class Solution:  # Similar to 3241. Time Taken to Mark All Nodes  def lastMarkedNodes(self, edges: list[list[int]]) -> list[int]:    n = len(edges) + 1    ans = [0] * n    tree = [[] for _ in range(n)]    # dp[i] := the last marked two nodes for subtree rooted at node i, where    # each node contains the time it got marked    dp = [Last2()] * n     for u, v in edges:      tree[u].append(v)      tree[v].append(u)     self._dfs(tree, 0, -1, dp)    self._reroot(tree, 0, -1, Node(), dp, ans)    return ans   def _dfs(      self,      tree: list[list[int]],      u: int,      prev: int,      dp: list[Last2]  ) -> 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.    """    last1 = Node(u, 0)    last2 = Node()    for v in tree[u]:      if v == prev:        continue      child = self._dfs(tree, v, u, dp)      time = child.time + 1      if time > last1.time:        last2 = last1        last1 = Node(child.node, time)      elif time > last2.time:        last2 = Node(child.node, time)    dp[u] = Last2(last1, last2)    return last1   def _reroot(      self,      tree: list[list[int]],      u: int,      prev: int,      last: Node,      dp: list[list[int]],      ans: list[int]  ) -> None:    """    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.    """    ans[u] = last.node if last.time > dp[u].last1.time else dp[u].last1.node    for v in tree[u]:      if v == prev:        continue      newLast = Node(last.node, last.time + 1)      if dp[u].last1.node == dp[v].last1.node:        alternativeTime = 1 + dp[u].last2.time        if alternativeTime > newLast.time:          newLast = Node(dp[u].last2.node, alternativeTime)      else:        alternativeTime = 1 + dp[u].last1.time        if alternativeTime > newLast.time:          newLast = Node(dp[u].last1.node, alternativeTime)      self._reroot(tree, v, u, newLast, dp, ans)

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