2673-make-costs-of-paths-equal-in-a-binary-tree.adoc

2673. Make Costs of Paths Equal in a Binary Tree

{leetcode}/problems/make-costs-of-paths-equal-in-a-binary-tree/[LeetCode - 2673. Make Costs of Paths Equal in a Binary Tree ^]

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.

Each node in the tree also has a cost represented by a given 0-indexed integer array cost of size n where cost[i] is the cost of node i + 1. You are allowed to increment the cost of any node by 1 any number of times.

Return the minimum number of increments you need to make the cost of paths from the root to each leaf node equal.

Note:

• A *perfect binary tree *is a tree where each node, except the leaf nodes, has exactly 2 children.

• The cost of a path is the sum of costs of nodes in the path.

Example 1: <img alt="" src="https://assets.leetcode.com/uploads/2023/04/04/binaryytreeedrawio-4.png" />

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.

Example 2: <img alt="" src="https://assets.leetcode.com/uploads/2023/04/04/binaryytreee2drawio.png" style="width: 205px; height: 151px;" />

Input: n = 3, cost = [5,3,3]
Output: 0
Explanation: 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

思路分析

一刷

link:{sourcedir}/_2673_MakeCostsOfPathsEqualInABinaryTree.java[role=include]

参考资料