Problem Description

You have a nested list structure where each element can be either:

• An integer
• A list containing more integers or nested lists

The depth of an integer is determined by how many levels of lists it's nested within. The outermost level has depth 1, elements inside one list have depth 2, elements inside two nested lists have depth 3, and so on.

For example, in the nested list [1,[2,2],[[3],2],1]:

• The first 1 and last 1 are at depth 1 (not inside any nested list)
• The two 2s in [2,2] are at depth 2 (inside one list)
• The 3 in [[3],2] is at depth 3 (inside two nested lists)
• The 2 in [[3],2] is at depth 2 (inside one list)

Your task is to calculate the weighted sum where each integer is multiplied by its depth, then sum all these products together.

For the example above: 1*1 + 2*2 + 2*2 + 3*3 + 2*2 + 1*1 = 1 + 4 + 4 + 9 + 4 + 1 = 23

The solution uses a depth-first search (DFS) approach to traverse the nested structure. Starting with depth 1, it recursively processes each element:

• If an element is an integer, multiply it by the current depth and add to the sum
• If an element is a list, recursively process that list with depth increased by 1

The NestedInteger interface provides methods to check if an element is an integer (isInteger()), get the integer value (getInteger()), or get the nested list (getList()).

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The nested list structure can be viewed as a graph where each list is a node that can contain integers (leaf nodes) or other lists (child nodes). This forms a tree-like hierarchical structure.

Is it a tree?

• Yes: The nested list structure is actually a tree. Each list can contain multiple elements (children), but there are no cycles - each element belongs to exactly one parent list. The structure has a clear root (the outermost list) and branches down to leaf nodes (integers).

DFS

• Yes: Since we identified this as a tree structure, the flowchart leads us directly to DFS (Depth-First Search).

Conclusion: The flowchart suggests using DFS for this problem, which makes perfect sense because:

1. We need to traverse each element in the nested structure
2. We need to track the depth/level as we go deeper into nested lists
3. DFS naturally maintains the depth information through its recursive call stack
4. Each integer's contribution to the sum depends on how deep it is in the structure

The DFS approach allows us to:

• Start at depth 1 for the outermost level
• Recursively explore nested lists, incrementing the depth
• Calculate the weighted sum by multiplying each integer by its current depth
• Aggregate all the weighted values as we traverse the entire structure

Intuition

When we look at this nested list structure, we can think of it like a file system where folders can contain files or other folders. Just as we might calculate the total size of files weighted by how deep they are in the directory structure, here we need to calculate a weighted sum based on nesting depth.

The key insight is recognizing that this is fundamentally a tree traversal problem. Each list is like a node that can have children (either integers or other lists). The "depth" mentioned in the problem is essentially the level of the node in this tree.

Why DFS? Consider how we naturally think about processing nested structures - we would:

1. Look at each element in the current list
2. If it's a number, we know its depth (current level) and can calculate its contribution
3. If it's another list, we need to "dive into" that list first before moving to the next element

This "dive deep first" approach is exactly what DFS does. The beautiful part about using recursion for DFS here is that the call stack naturally keeps track of our current depth. Each recursive call represents going one level deeper, so we can simply pass depth + 1 to the recursive call.

Think about manually calculating the sum for [1,[2,2],[[3],2],1]:

• Start at depth 1: see 1 (contribute 1*1)
• Still at depth 1: see [2,2] - need to go deeper
  • Now at depth 2: see 2 (contribute 2*2), see another 2 (contribute 2*2)
• Back to depth 1: see [[3],2] - need to go deeper
  • Now at depth 2: see [3] - need to go even deeper
    • Now at depth 3: see 3 (contribute 3*3)
  • Back to depth 2: see 2 (contribute 2*2)
• Back to depth 1: see 1 (contribute 1*1)

This natural way of thinking through the problem maps perfectly to a recursive DFS solution where each function call handles one level of nesting and returns the sum of all integers at that level and below.

Solution Approach

The implementation uses a recursive DFS helper function to traverse the nested list structure. Let's walk through the solution step by step:

Main Function Setup:

def depthSum(self, nestedList: List[NestedInteger]) -> int:
    return dfs(nestedList, 1)

We start by calling our helper function with the initial list and depth of 1.

DFS Helper Function:

def dfs(nestedList, depth):
    depth_sum = 0
    for item in nestedList:
        if item.isInteger():
            depth_sum += item.getInteger() * depth
        else:
            depth_sum += dfs(item.getList(), depth + 1)
    return depth_sum

The algorithm works as follows:

1. Initialize a local sum: For each recursive call, we maintain depth_sum to accumulate the weighted values at this level and all deeper levels.

2. Iterate through each element: We process each NestedInteger object in the current list.

3. Check element type: Using the isInteger() method from the NestedInteger interface:

   • If it's an integer: We get its value using getInteger(), multiply by the current depth, and add to our sum
   • If it's a nested list: We recursively call dfs() with:
     • The nested list obtained via getList()
     • Incremented depth (depth + 1)
     • Add the returned sum to our current depth_sum

4. Return the accumulated sum: After processing all elements at this level, return the total sum.

Key Design Decisions:

• Recursion over iteration: While we could use an iterative approach with a stack, recursion naturally handles the depth tracking through the call stack.

• Passing depth as parameter: Instead of maintaining a global depth variable, we pass depth as a parameter, making the function pure and easier to reason about.

• Single pass traversal: Each element is visited exactly once, making the solution efficient with O(n) time complexity where n is the total number of integers in the structure.

Example Walkthrough with [1,[2,2],[[3],2],1]:

• dfs([1,[2,2],[[3],2],1], 1) starts
  • Processes 1: adds 1*1 = 1
  • Processes [2,2]: calls dfs([2,2], 2)
    • Adds 2*2 = 4 and 2*2 = 4, returns 8
  • Processes [[3],2]: calls dfs([[3],2], 2)
    • Processes [3]: calls dfs([3], 3)
      • Adds 3*3 = 9, returns 9
    • Processes 2: adds 2*2 = 4
    • Returns 13
  • Processes 1: adds 1*1 = 1
  • Returns total: 1 + 8 + 13 + 1 = 23

Example Walkthrough

Let's walk through a simpler example: [[1,1],2,[1,1]]

Visual representation of the structure:

[
  [1, 1],    // depth 1: this is a list
  2,         // depth 1: integer value 2
  [1, 1]     // depth 1: this is a list
]

Step-by-step execution:

1. Start: dfs([[1,1],2,[1,1]], depth=1)

   • Initialize depth_sum = 0
   • Begin iterating through the list

2. First element [1,1]:

   • It's not an integer (it's a list)
   • Make recursive call: dfs([1,1], depth=2)
     • In this recursive call with depth=2:
       • First 1: is integer → add 1 * 2 = 2
       • Second 1: is integer → add 1 * 2 = 2
       • Return 4 back to parent call
   • Add returned value: depth_sum = 0 + 4 = 4

3. Second element 2:

   • It's an integer
   • Calculate: 2 * 1 = 2 (depth is 1)
   • Add to sum: depth_sum = 4 + 2 = 6

4. Third element [1,1]:

   • It's not an integer (it's a list)
   • Make recursive call: dfs([1,1], depth=2)
     • In this recursive call with depth=2:
       • First 1: is integer → add 1 * 2 = 2
       • Second 1: is integer → add 1 * 2 = 2
       • Return 4 back to parent call
   • Add returned value: depth_sum = 6 + 4 = 10

5. Final result: Return 10

Calculation breakdown:

• Two 1s from first [1,1] at depth 2: 1*2 + 1*2 = 4
• One 2 at depth 1: 2*1 = 2
• Two 1s from second [1,1] at depth 2: 1*2 + 1*2 = 4
• Total: 4 + 2 + 4 = 10

This example demonstrates how the recursive DFS naturally handles the depth tracking - each time we encounter a nested list, we increment the depth by 1 in the recursive call, and when we return from that call, we're automatically back at the previous depth level.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the total number of elements (including nested elements) in the entire nested list structure.

The algorithm uses depth-first search (DFS) to traverse through all elements in the nested list. Each element is visited exactly once:

• For integers: We perform a constant time operation (getInteger() and multiplication)
• For nested lists: We make a recursive call to process its contents

Since every element (whether it's an integer or a nested list) is processed exactly once, the time complexity is linear with respect to the total number of elements.

Space Complexity: O(d) where d is the maximum depth of nesting in the input.

The space complexity is determined by:

• The recursive call stack: In the worst case, the recursion depth equals the maximum nesting level of the input structure
• Local variables in each recursive call: Only a constant amount of space (depth_sum, item, depth) is used per recursive call

For example, for a deeply nested structure like [[[[1]]]], the recursion would go 4 levels deep, using O(4) space on the call stack. The space complexity does not depend on the total number of elements but rather on how deeply they are nested.

Common Pitfalls

1. Incorrect Depth Initialization

One of the most common mistakes is starting the depth at 0 instead of 1. The problem clearly states that the outermost level has depth 1, not depth 0.

Incorrect Implementation:

def depthSum(self, nestedList: List[NestedInteger]) -> int:
    return calculate_depth_sum(nestedList, 0)  # Wrong! Starting at depth 0

Why it's wrong: This would calculate 1*0 + 2*1 + 2*1 + 3*2 + 2*1 + 1*0 = 0 + 2 + 2 + 6 + 2 + 0 = 12 instead of the correct answer 23.

Solution: Always start with depth 1 as shown in the correct implementation.

2. Mishandling the NestedInteger Interface

Another pitfall is incorrectly using the NestedInteger methods or making assumptions about the data structure.

Common Mistakes:

# Mistake 1: Trying to iterate directly over a NestedInteger
for item in element:  # Wrong! NestedInteger is not iterable
    ...

# Mistake 2: Forgetting to call getList() for nested lists
if not element.isInteger():
    total_sum += calculate_depth_sum(element, depth + 1)  # Wrong! Should be element.getList()

# Mistake 3: Assuming getInteger() works on lists
value = element.getInteger()  # This returns None if element is a list!

Solution: Always check isInteger() first, then use the appropriate method:

• Use getInteger() only when isInteger() returns True
• Use getList() only when isInteger() returns False

3. Incrementing Depth Incorrectly

Some implementations mistakenly modify the depth variable instead of passing depth + 1.

Incorrect Implementation:

def calculate_depth_sum(nested_list, depth):
    total_sum = 0
    for element in nested_list:
        if element.isInteger():
            total_sum += element.getInteger() * depth
        else:
            depth += 1  # Wrong! This affects subsequent elements at the same level
            total_sum += calculate_depth_sum(element.getList(), depth)
    return total_sum

Why it's wrong: If you have [1, [2], 3], after processing [2], the depth would be permanently increased, causing 3 to be calculated with the wrong depth.

Solution: Always pass depth + 1 as a parameter without modifying the current depth variable:

total_sum += calculate_depth_sum(element.getList(), depth + 1)

4. Edge Case: Empty Lists

Not handling empty nested lists properly can lead to issues in some implementations.

Potential Issue:

# Some might try to optimize by checking list length first
if len(element.getList()) > 0:  # Unnecessary and could cause issues
    total_sum += calculate_depth_sum(element.getList(), depth + 1)

Solution: The recursive function naturally handles empty lists by returning 0 (the loop simply doesn't execute), so no special handling is needed. The implementation should work correctly with inputs like [[],[[]],1].

5. Type Confusion with Mixed Implementations

When solving this problem outside of LeetCode's environment, developers might mix different representations.

Common Mistake:

# Mixing Python lists with NestedInteger objects
def calculate_depth_sum(nested_list, depth):
    total_sum = 0
    for element in nested_list:
        if isinstance(element, int):  # Wrong! Should use isInteger()
            total_sum += element * depth
        elif isinstance(element, list):  # Wrong! Should use getList()
            total_sum += calculate_depth_sum(element, depth + 1)
    return total_sum

Solution: Consistently use the NestedInteger interface methods (isInteger(), getInteger(), getList()) rather than Python's built-in type checking.