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Problem solution · Python

Most Profitable Path in a Tree

Most Profitable Path in a 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
60 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 Most Profitable Path in 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

  • 60 lines of Python from the credited upstream file 2467.py.
  • The implementation visibly relies on sequence storage.
  • 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 codeMost Profitable Path in a Tree · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def mostProfitablePath(      self,      edges: list[list[int]],      bob: int,      amount: list[int],  ) -> int:    n = len(amount)    tree = [[] for _ in range(n)]    parent = [0] * n    aliceDist = [-1] * n     for u, v in edges:      tree[u].append(v)      tree[v].append(u)     # Fills `parent` and `aliceDist`.    def dfs(u: int, prev: int, d: int) -> None:      parent[u] = prev      aliceDist[u] = d      for v in tree[u]:        if aliceDist[v] == -1:          dfs(v, u, d + 1)     dfs(0, -1, 0)     # Modify amount athe path from node bob to node 0.    # For each node,    #   1. If Bob reaches earlier than Alice does, change the amount to 0.    #   2. If Bob and Alice reach simultaneously, devide the amount by 2.    u = bob    bobDist = 0    while u != 0:      if bobDist < aliceDist[u]:        amount[u] = 0      elif bobDist == aliceDist[u]:        amount[u] //= 2      u = parent[u]      bobDist += 1     return self._getMoney(tree, 0, -1, amount)   def _getMoney(      self,      tree: list[list[int]],      u: int,      prev: int,      amount: list[int],  ) -> int:    # a leaf node    if len(tree[u]) == 1 and tree[u][0] == prev:      return amount[u]     maxPath = -math.inf    for v in tree[u]:      if v != prev:        maxPath = max(maxPath, self._getMoney(tree, v, u, amount))     return amount[u] + maxPath

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