Leetcode: 339. Nested List Weight Sum

Given a nested list of integers, return the sum of all integers in the list weighted by their depth.

Each element is either an integer, or a list -- whose elements may also be integers or other lists.

Example 1:
Given the list [[1,1],2,[1,1]], return 10. (four 1's at depth 2, one 2 at depth 1)

Example 2:
Given the list [1,[4,[6]]], return 27. (one 1 at depth 1, one 4 at depth 2, and one 6 at depth 3; 1 + 4*2 + 6*3 = 27)

Understand the problem:
Because the input is nested, it is natural to think about the problem in a recursive way. We go through the list of nested integers one by one, keeping track of the current depth d. If a nested integer is an integer n, we calculate its sum as n * d. If the nested integer is a list, we calculate the sum of this list recursively using the same process but with depth d+1.

Code (Java):

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * public interface NestedInteger {
 *
 *     // @return true if this NestedInteger holds a single integer, rather than a nested list.
 *     public boolean isInteger();
 *
 *     // @return the single integer that this NestedInteger holds, if it holds a single integer
 *     // Return null if this NestedInteger holds a nested list
 *     public Integer getInteger();
 *
 *     // @return the nested list that this NestedInteger holds, if it holds a nested list
 *     // Return null if this NestedInteger holds a single integer
 *     public List<NestedInteger> getList();
 * }
 */
public class Solution {
    public int depthSum(List<NestedInteger> nestedList) {
        if (nestedList == null || nestedList.size() == 0) {
            return 0;
        }
        
        return depthSumHelper(nestedList.iterator(), 1);
    }
    
    private int depthSumHelper(Iterator<NestedInteger> iterator, int depth) {
        if (!iterator.hasNext()) {
            return 0;
        }
        
        int sum = 0;
        
        while (iterator.hasNext()) {
            NestedInteger curr = iterator.next();
            if (curr.isInteger()) {
                sum += curr.getInteger() * depth;
            } else {
                sum += depthSumHelper(curr.getList().iterator(), depth + 1);
            }
        }
        
        return sum;
    }
}

Another solution without passing an iterator:

public int depthSum(List<NestedInteger> nestedList) {
    return depthSum(nestedList, 1);
}

public int depthSum(List<NestedInteger> list, int depth) {
    int sum = 0;
    for (NestedInteger n : list) {
        if (n.isInteger()) {
            sum += n.getInteger() * depth;
        } else {
            sum += depthSum(n.getList(), depth + 1);
        }
    }
    return sum;
}

Analysis:

The algorithm takes $O(N)$ time, where $N$ is the total number of nested elements in the input list. For example, the list [ [[[[1]]]], 2 ] contains $4$ nested lists and $2$ nested integers ($1$ and $2$), so $N=6$.

In terms of space, at most $O(D)$ recursive calls are placed on the stack, where $D$ is the maximum level of nesting in the input. For example, $D=2$ for the input [[1,1],2,[1,1]], and $D=3$ for the input [1,[4,[6]]].