#2673

Medium

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Make Costs of Paths Equal in a Binary Tree

Minimize the cost increments required to equalize path costs in a binary tree from root to leaves.

array dynamic programming greedy tree binary tree

Problem Statement

You are given an integer n representing the number of nodes in a perfect binary tree. The tree consists of nodes numbered from 1 to n, with the root being node 1. Each node i has two children, where the left child is node 2 * i and the right child is node 2 * i + 1. You are also given a 0-indexed integer array cost where cost[i] represents the cost of node i + 1 in the tree.

The objective is to return the minimum number of increments needed to make the cost of the paths from the root to each leaf node equal. You can increment the cost of any node by 1 any number of times. You are required to calculate this minimum increment count efficiently.

Examples

Example 1

Input: n = 7, cost = [1,5,2,2,3,3,1]

Output: 6

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.

Example 2

Input: n = 3, cost = [5,3,3]

Output: 0

The two paths already have equal total costs, so no increments are needed.

Constraints

3 <= n <= 105
n + 1 is a power of 2
cost.length == n
1 <= cost[i] <= 104

Solution Approach

Binary Tree Traversal

The solution starts with a traversal of the binary tree, exploring each path from the root to all leaf nodes. For each path, track the sum of node costs and identify the path with the highest cost.

State Tracking and Increment Calculation

Once the path with the maximum cost is identified, calculate how many increments are needed for each other path to match this maximum cost. This involves comparing the total path cost of each leaf node with the highest cost path.

Dynamic Programming / Greedy Approach

A dynamic programming or greedy approach is applied to determine the minimal number of increments required. This can be achieved by efficiently adjusting the costs using state tracking, ensuring the solution remains optimal in terms of time complexity.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the traversal method used (DFS/BFS) and the way the cost comparison and increment calculation is handled. It could range from O(n) to O(n log n), depending on the optimization approach used, especially for larger trees with n up to 10^5. The space complexity is also tied to the tree structure and the traversal technique, typically O(n).

What Interviewers Usually Probe

Look for a clear understanding of binary tree traversal techniques (DFS/BFS).
Evaluate the candidate's ability to optimize the solution using dynamic programming or greedy strategies.
Assess how well the candidate balances time complexity and correctness, particularly for large inputs.

Common Pitfalls or Variants

Common pitfalls

Forgetting to adjust node costs optimally, leading to unnecessary increments.
Not recognizing the importance of path cost comparison between the highest and others.
Overcomplicating the solution without leveraging greedy or dynamic programming techniques effectively.

Follow-up variants

Adjusting the tree structure by introducing non-perfect binary trees.
Incorporating constraints where cost increases are limited to a fixed range.
Extending the problem to multiple trees with varying costs and structure.

FAQ

What is the primary approach to solving 'Make Costs of Paths Equal in a Binary Tree'?

The problem is solved using binary tree traversal and state tracking, applying either dynamic programming or greedy strategies to minimize the increments.

How do I ensure minimal increments in the solution?

To minimize increments, identify the path with the maximum cost and adjust other paths to match it by incrementing node costs as necessary.

How does the structure of the binary tree affect the solution?

The perfect binary tree structure ensures that each path has the same number of nodes, simplifying the path comparison for cost adjustments.

What are the common mistakes in solving this problem?

Common mistakes include incorrect path cost comparison and inefficient handling of the increment calculations, leading to higher time complexity.

Can this problem be solved without dynamic programming or greedy techniques?

While possible, solving the problem efficiently without dynamic programming or greedy techniques would likely result in a less optimal solution in terms of time complexity.

terminal

Solution

Solution 1: Greedy Algorithm

According to the problem description, we need to calculate the minimum number of increments to make the path values from the root node to each leaf node equal.

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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

Continue Practicing

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