Make Costs of Paths Equal in a Binary Tree leetcode solution

 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:

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

solution:

class Solution { public:   int ansi(int n  , vector<int>&cost , int idx , int &ans )       {             // int  =cost[0];             if(2*idx +2 >= n && 2*idx+1 >= n)               return cost[idx];             int a = ansi(n , cost , 2*idx+2 ,ans )  ;             int b = ansi(n , cost ,2*idx+1 ,ans);             // int mini = a < b ? a : b;             // int maxi = a> b ? a : b ;             ans += abs(a -b);         return cost[idx] + max(a, b);       }     int minIncrements(int n, vector<int>& cost) {               int ans = 0;         int idx =0;         ansi(n, cost , idx , ans);         return ans;     } };