Make Costs of Paths Equal in a Binary Tree — LeetCode 2673 Python Solution

The problem

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.

Example

Input

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

Output

6

Explanation

We can do the following increments:

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

Pattern: Tree Traversal

Choose the order — preorder, inorder, postorder, level — and the problem solves itself. LeetCode 2673. Make Costs of Paths Equal in a Binary Tree is filed here because LeetCode tags it Tree and Binary Tree, which is the vocabulary this hub collects.

The tree traversal guide has the Python template for the pattern and the 225 LeetCode problems that use it.

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