LeetCode每日一题2673. Make Costs of Paths Equal in a Binary Tree

一、题目

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:

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:

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

二、题解

class Solution {
public:int minIncrements(int n, vector<int>& cost) {int res = 0;for(int i = n / 2;i > 0;i--){res += abs(cost[i * 2 - 1] - cost[i * 2]);cost[i - 1] += max(cost[i * 2 - 1],cost[i * 2]);}return res;}
};