Leetcode #2673: Make Costs of Paths Equal in a Binary Tree
In this guide, we solve Leetcode #2673 Make Costs of Paths Equal in a Binary 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
You are given an integer n representing the number of nodes in a perfect binary tree consisting of nodes numbered from 1 to n. The root of the tree is node 1 and each node i in the tree has two children where the left child is the node 2 * i and the right child is 2 * i + 1.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Tree, Array, Dynamic Programming, Binary Tree
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: n = 7, cost = [1,5,2,2,3,3,1]
Output: 6
Explanation: We can do the following increments:
- Increase the cost of node 4 one time.
- Increase the cost of node 3 three times.
- Increase the cost of node 7 two times.
Each path from the root to a leaf will have a total cost of 9.
The total increments we did is 1 + 3 + 2 = 6.
It can be shown that this is the minimum answer we can achieve.
Python Solution
class Solution:
def minIncrements(self, n: int, cost: List[int]) -> int:
ans = 0
for i in range(n >> 1, 0, -1):
l, r = i << 1, i << 1 | 1
ans += abs(cost[l - 1] - cost[r - 1])
cost[i - 1] += max(cost[l - 1], cost[r - 1])
return ans
Complexity
The time complexity is , where is the number of nodes. 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.