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Problem solution · Python

Time Taken to Mark All Nodes

Time Taken to Mark All Nodes: 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
94 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 Time Taken to Mark All Nodes, 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

  • 94 lines of Python from the credited upstream file 3241.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 codeTime Taken to Mark All Nodes · 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 taken to mark the entire subtree rooted at the node  class Top2:  def __init__(self, top1: Node = Node(), top2: Node = Node()):    # the direct child node, where the time taken to mark the entire subtree    # rooted at the node is the maximum    self.top1 = top1    # the direct child node, where the time taken to mark the entire subtree    # rooted at the node is the second maximum    self.top2 = top2  class Solution:  def timeTaken(self, edges: list[list[int]]) -> list[int]:    n = len(edges) + 1    ans = [0] * n    tree = [[] for _ in range(n)]    # dp[i] := the top two direct child nodes for subtree rooted at node i,    # where each node contains the time taken to mark the entire subtree rooted    # at the node itself    dp = [Top2()] * 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, 0, dp, ans)    return ans   def _getTime(self, u: int) -> int:    """Returns the time taken to mark node u."""    return 2 if u % 2 == 0 else 1   def _dfs(      self,      tree: list[list[int]],      u: int,      prev: int,      dp: list[Top2]  ) -> int:    """    Performs a DFS traversal of the subtree rooted at node `u`, computes the    time taken to mark all nodes in the subtree, records the top two direct    child nodes, where the time taken to mark the subtree rooted at each of the    child nodes is maximized, and returns the top child node.     These values are used later in the rerooting process.    """    top1 = Node()    top2 = Node()    for v in tree[u]:      if v == prev:        continue      time = self._dfs(tree, v, u, dp) + self._getTime(v)      if time >= top1.time:        top2 = top1        top1 = Node(v, time)      elif time > top2.time:        top2 = Node(v, time)    dp[u] = Top2(top1, top2)    return top1.time   def _reroot(      self,      tree: list[list[int]],      u: int,      prev: int,      maxTime: int,      dp: list[Top2],      ans: list[int]  ) -> None:    """    Reroots the tree at node `u` and updates the answer array, where `maxTime`    is the longest path that doesn't go through `u`'s subtree.    """    ans[u] = max(maxTime, dp[u].top1.time)     for v in tree[u]:      if v == prev:        continue      newMaxTime = self._getTime(u) + max(          maxTime,          dp[u].top2.time if dp[u].top1.node == v else dp[u].top1.time      )      self._reroot(tree, v, u, newMaxTime, dp, ans)

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