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2673. Make Costs of Paths Equal in a Binary Tree

Description

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

Solutions

  • class Solution {
    private int[] cost;
    private int n;
    private int ans;

    public int minIncrements(int n, int[] cost) {
        this.n = n;
        this.cost = cost;
        dfs(1);
        return ans;
    }

    private int dfs(int i) {
        if ((i << 1) > n) {
            return cost[i - 1];
        }
        int l = dfs(i << 1);
        int r = dfs(i << 1 | 1);
        ans += Math.max(l, r) - Math.min(l, r);
        return cost[i - 1] + Math.max(l, r);
    }
}

  • class Solution {
public:
    int minIncrements(int n, vector<int>& cost) {
        int ans = 0;
        function<int(int)> dfs = [&](int i) -> int {
            if ((i << 1) > n) {
                return cost[i - 1];
            }
            int l = dfs(i << 1);
            int r = dfs(i << 1 | 1);
            ans += max(l, r) - min(l, r);
            return cost[i - 1] + max(l, r);
        };
        dfs(1);
        return ans;
    }
};

  • class Solution:
    def minIncrements(self, n: int, cost: List[int]) -> int:
        def dfs(i: int) -> int:
            if (i << 1) > n:
                return cost[i - 1]
            l, r = dfs(i << 1), dfs(i << 1 | 1)
            nonlocal ans
            ans += max(l, r) - min(l, r)
            return cost[i - 1] + max(l, r)

        ans = 0
        dfs(1)
        return ans


  • func minIncrements(n int, cost []int) (ans int) {
	var dfs func(int) int
	dfs = func(i int) int {
		if (i << 1) > n {
			return cost[i-1]
		}
		l, r := dfs(i<<1), dfs(i<<1|1)
		ans += max(l, r) - min(l, r)
		return cost[i-1] + max(l, r)
	}
	dfs(1)
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

  • function minIncrements(n: number, cost: number[]): number {
    let ans = 0;
    const dfs = (i: number): number => {
        if (i << 1 > n) {
            return cost[i - 1];
        }
        const [a, b] = [dfs(i << 1), dfs((i << 1) | 1)];
        ans += Math.max(a, b) - Math.min(a, b);
        return cost[i - 1] + Math.max(a, b);
    };
    dfs(1);
    return ans;
}

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