Leetcode #2467: Most Profitable Path in a Tree
In this guide, we solve Leetcode #2467 Most Profitable Path in a Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is an undirected tree with n nodes labeled from 0 to n - 1, rooted at node 0. You are given a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Graph, Array
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: edges = [[0,1],[1,2],[1,3],[3,4]], bob = 3, amount = [-2,4,2,-4,6]
Output: 6
Explanation:
The above diagram represents the given tree. The game goes as follows:
- Alice is initially on node 0, Bob on node 3. They open the gates of their respective nodes.
Alice's net income is now -2.
- Both Alice and Bob move to node 1.
Since they reach here simultaneously, they open the gate together and share the reward.
Alice's net income becomes -2 + (4 / 2) = 0.
- Alice moves on to node 3. Since Bob already opened its gate, Alice's income remains unchanged.
Bob moves on to node 0, and stops moving.
- Alice moves on to node 4 and opens the gate there. Her net income becomes 0 + 6 = 6.
Now, neither Alice nor Bob can make any further moves, and the game ends.
It is not possible for Alice to get a higher net income.
Python Solution
class Solution:
def mostProfitablePath(
self, edges: List[List[int]], bob: int, amount: List[int]
) -> int:
def dfs1(i, fa, t):
if i == 0:
ts[i] = min(ts[i], t)
return True
for j in g[i]:
if j != fa and dfs1(j, i, t + 1):
ts[j] = min(ts[j], t + 1)
return True
return False
def dfs2(i, fa, t, v):
if t == ts[i]:
v += amount[i] // 2
elif t < ts[i]:
v += amount[i]
nonlocal ans
if len(g[i]) == 1 and g[i][0] == fa:
ans = max(ans, v)
return
for j in g[i]:
if j != fa:
dfs2(j, i, t + 1, v)
n = len(edges) + 1
g = defaultdict(list)
ts = [n] * n
for a, b in edges:
g[a].append(b)
g[b].append(a)
dfs1(bob, -1, 0)
ts[bob] = 0
ans = -inf
dfs2(0, -1, 0, 0)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.