Make Costs of Paths Equal in a Binary Tree | HARD | TRICKY | MUST DO | OA

https://leetcode.com/problems/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:
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 <= 10^5
n + 1 is a power of 2
cost.length == n
1 <= cost[i] <= 10^4

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class Solution {
public:
    int res=0;

    int dfs(vector<int> &cost, int node){
        if(node>cost.size()){
            return 0;
        }
        int L=dfs(cost,2*node);
        int R=dfs(cost,2*node+1);
        res+=abs(L-R);
        return cost[node-1]+max(L,R);
    }

    int minIncrements(int n, vector<int>& cost) {
        dfs(cost,1);
        return res;
    }
};