LeetCode 2673: Make Costs of Paths Equal in a Binary Tree Solution

Master LeetCode problem 2673 (Make Costs of Paths Equal in a Binary Tree), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2673. Make Costs of Paths Equal in a Binary Tree

Medium

Array Dynamic Programming Greedy Tree Binary Tree

Problem Explanation

Explanation:

To make the cost of paths from the root to each leaf node equal, we need to balance the costs in the tree. We can achieve this by iteratively adjusting the costs of parent nodes based on the costs of their children. The idea is to propagate the excess costs from children to their parents such that the total cost of each path from the root to a leaf becomes equal. We will perform a depth-first traversal of the tree while keeping track of the excess cost at each node.

Algorithm:

Perform a depth-first traversal of the tree from the root.
Calculate the excess cost for each child node by subtracting the average cost of its children from its own cost.
Propagate the excess cost to the parent node by adding the excess cost of its children.
Recursively repeat steps 2 and 3 for each child node.

Time Complexity: The time complexity of this algorithm is O(n) where n is the number of nodes in the binary tree.

Space Complexity: The space complexity of this algorithm is O(n) for the recursive stack.

Solution Code

class Solution {
    public int minCost(int n, int[] cost) {
        return dfs(0, 0, cost);
    }
    
    private int dfs(int idx, int depth, int[] cost) {
        if (idx >= cost.length) return 0;
        
        int left = dfs(2 * idx + 1, depth + 1, cost);
        int right = dfs(2 * idx + 2, depth + 1, cost);
        
        int totalCost = left + right + cost[idx];
        
        if (depth > 0) {
            int excess = totalCost - cost[idx];
            cost[idx] += excess;
            return excess;
        }
        
        return totalCost;
    }
}

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