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.
\n\nNote:
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Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2023/04/04/binaryytreeedrawio-4.png\")
\n\nInput: n = 7, cost = [1,5,2,2,3,3,1]\nOutput: 6\nExplanation: We can do the following increments:\n- Increase the cost of node 4 one time.\n- Increase the cost of node 3 three times.\n- Increase the cost of node 7 two times.\nEach path from the root to a leaf will have a total cost of 9.\nThe total increments we did is 1 + 3 + 2 = 6.\nIt can be shown that this is the minimum answer we can achieve.\n\n\n
Example 2:
\n![\"\"](\"https://assets.leetcode.com/uploads/2023/04/04/binaryytreee2drawio.png\")
\n\nInput: n = 3, cost = [5,3,3]\nOutput: 0\nExplanation: The two paths already have equal total costs, so no increments are needed.\n\n\n
\n
Constraints:
\n\n3 <= n <= 105n + 1 is a power of 2cost.length == n1 <= cost[i] <= 104Compiled with clang 11 using the latest C++ 20 standard.
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