2673. Make Costs of Paths Equal in a Binary Tree

class Solution {
    public int minIncrements(int n, int[] cost) {
        //   *        0
        //  * *      1  2
        // * * *   3 4 5 6

        int minIncrements = 0;

        // 倒序遍历非叶子节点
        for (int i = cost.length / 2 - 1; i >= 0; i--) {
            minIncrements += Math.abs(cost[i * 2 + 1] - cost[i * 2 + 2]); // 将较小的子节点的值增大
            cost[i] += Math.max(cost[i * 2 + 1], cost[i * 2 + 2]); // 将较大的子节点的值累加至父节点，以使更上一层时只需参考下一层的节点值
        }

        return minIncrements;
    }
}

References

2673. Make Costs of Paths Equal in a Binary Tree