Delete Tree Nodes

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TreesRecursion

A tree is rooted at node 0 consisting of n nodes numbered from 0 to n - 1. You are given a 0-indexed integer array values of length n, where values[i] is the value of the i^th node. You are also given a 0-indexed 2D integer array edges of size (n - 1) x 2, where edges[i] = [u_i, v_i] means there is an undirected edge between nodes u_i and v_i.

Consider performing the following operation until you are left with at most one node:

Choose a node that has at most one neighbor (i.e., it is a leaf or a root).
Delete it along with all its incident edges. The values of the remaining nodes remain unchanged.

Return the maximum possible sum of the values of the nodes remaining after performing the operation optimally.

Example 1:

Input: values = [-1,-2,-3,-4,2], edges = [[0,1],[1,2],[2,3],[3,4]]
Output: 2
Explanation: Perform the following operations:
1. Delete node 0. The tree now consists of nodes [1,2,3,4].
2. Delete node 1. The tree now consists of nodes [2,3,4].
3. Delete node 3. The tree now consists of nodes [2,4].
4. Delete node 2. The tree now consists of node [4].
There is only one node left, which is node 4, so the answer is values[4] = 2.

Example 2:

Input: values = [2,4,8,16], edges = [[0,1],[1,2],[1,3]]
Output: 30
Explanation: Perform the following operations:
1. Delete node 0. The tree now consists of nodes [1,2,3].
2. Delete node 2. The tree now consists of nodes [1,3].
After the 2nd operation, only nodes 1 and 3 are left, so the answer is values[1] + values[3] = 24.

Constraints:

1 <= n <= 2 * 10⁴
values.length == n
-10⁹ <= values[i] <= 10⁹
0 <= edges.length < n
edges[i].length == 2
0 <= u_i, v_i < n
The input is generated such that edges represents a valid tree.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for `nodes`, the values in the `value` array, and the elements in the `parent` array?
Is the input guaranteed to represent a valid tree structure (e.g., is there always exactly one root, and are there no cycles other than self-loops on the root)?
If the entire tree sums to zero, should I return 0, or is there some other special value I should return?
Can a node have a value of zero initially, and does a zero-valued node automatically cause its subtree to be deleted if the rest of the subtree sums to zero?
Are the node IDs guaranteed to be consecutive, starting from 0 up to `nodes` - 1?

Brute Force Solution

Approach

The brute force approach to deleting tree nodes means we are going to try removing every possible combination of nodes and then checking the resulting tree. We'll keep track of which combination results in the best outcome according to the problem's requirements.

Here's how the algorithm would work step-by-step:

First, consider the scenario where we delete no nodes at all. This is our starting point.
Next, try deleting only the first node, and then see if the resulting tree meets the problem's criteria.
Then, try deleting only the second node, and again check if it meets the criteria.
Repeat this for every single node in the tree, one at a time.
Now, try deleting pairs of nodes. Take the first and second node, then the first and third node, and so on, checking the tree each time.
Continue this process, trying to delete all possible combinations of three nodes, four nodes, and so on, up to deleting all nodes except one.
For each possible tree (after deleting some nodes), evaluate its 'score' based on the problem's rules.
After checking all possibilities, pick the tree with the best score according to the problem's instructions. This is your answer.

Code Implementation

def delete_tree_nodes_brute_force(parent, values):
    number_of_nodes = len(parent)
    best_nodes = list(range(number_of_nodes))
    
    for i in range(1 << number_of_nodes):
        nodes_to_delete = []
        for j in range(number_of_nodes):
            if (i >> j) & 1:
                nodes_to_delete.append(j)

        remaining_nodes = []
        for node_index in range(number_of_nodes):
            if node_index not in nodes_to_delete:
                remaining_nodes.append(node_index)

        # Create a new parent array representing the remaining tree
        new_parent = [-1] * len(remaining_nodes)
        node_map = {node_index: index for index, node_index in enumerate(remaining_nodes)}
        is_valid = True

        for node_index in remaining_nodes:
            original_parent = parent[node_index]
            if original_parent != -1:
                if original_parent in remaining_nodes:
                    new_parent[node_map[node_index]] = node_map[original_parent]
                elif original_parent not in remaining_nodes:
                    # Parent was deleted
                    continue
            else:
                new_parent[node_map[node_index]] = -1

        #Check if it is a better solution.
        if is_valid:
            if len(remaining_nodes) > len(best_nodes):
                best_nodes = remaining_nodes
            elif len(remaining_nodes) == len(best_nodes):
                remaining_sum = sum(values[node_index] for node_index in remaining_nodes)
                best_sum = sum(values[node_index] for node_index in best_nodes)

                #Prefer solutions with smaller sum
                if remaining_sum < best_sum:
                    best_nodes = remaining_nodes
    
    #We only want nodes that are not deleted
    return best_nodes

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible combinations of deleting nodes from the tree. Given n nodes, there are 2^n possible subsets of nodes that can be deleted (each node is either deleted or not). For each subset, the algorithm needs to potentially reconstruct and evaluate the resulting tree. This evaluation involves at least visiting the remaining nodes, but the dominant factor is the enumeration of all subsets, leading to a time complexity of O(2^n).

Space Complexity

O(1) – The brute force approach, as described, does not employ auxiliary data structures that scale with the input size N (number of nodes in the tree). It explores different combinations by iteratively considering nodes for deletion, but it doesn't explicitly store intermediate trees or node combinations in a way that significantly increases memory usage. The primary operation is checking if a modified tree meets certain criteria; this likely involves a constant amount of extra space for calculations within the checking function. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The trick to solving this problem efficiently is to start from the bottom of the tree and work our way up. We need to figure out which nodes to remove *before* we start deleting anything, so we don't mess up the tree structure prematurely.

Here's how the algorithm would work step-by-step:

Imagine starting at the very bottom of the tree, at the leaf nodes.
For each node, check if its value needs to be removed based on the problem's rules (e.g., if the sum of its value and its children's values is zero).
If a node should be removed, mark it for deletion. But don't delete it yet!
Move up to the node's parent.
Repeat the checking and marking process for the parent node, including the already-marked children's values in the calculation.
Keep moving up the tree, one level at a time, until you reach the root node, marking nodes for deletion as needed.
Once you've reached the root and marked all the nodes that should be removed, *then* go through the tree and actually delete the marked nodes.
When deleting a node, remember to also remove the connection from its parent to it.
This approach ensures that you always have the correct tree structure to calculate whether a node should be deleted, without accidentally cutting things off early.

Code Implementation

def delete_tree_nodes(number_of_nodes, parent_nodes, node_values):

    children = [[] for _ in range(number_of_nodes)]
    for node_index in range(1, number_of_nodes):
        children[parent_nodes[node_index]].append(node_index)

    nodes_to_keep = set(range(number_of_nodes))

    def calculate_sum(node_index):
        if node_index not in nodes_to_keep:
            return 0

        current_sum = node_values[node_index]
        for child_index in children[node_index]:
            current_sum += calculate_sum(child_index)

        # If the sum is zero, mark node for deletion.
        if current_sum == 0:
            nodes_to_keep.discard(node_index)

        return current_sum

    calculate_sum(0)
    # Only count nodes that remain after deletion
    return len(nodes_to_keep)

Big(O) Analysis

Time Complexity

O(n) – The algorithm performs a depth-first traversal of the tree, visiting each of the n nodes once to determine whether it should be deleted. During this traversal, a constant amount of work is done at each node (checking its value and potentially updating the parent's reference). After the traversal, there could be another pass to physically delete the nodes, but each of those nodes is visited once and only once to be removed. Therefore the runtime is proportional to the number of nodes, resulting in O(n) time complexity.

Space Complexity

O(N) – The algorithm implicitly uses a recursion stack, and in the worst-case scenario (e.g., a skewed tree), the depth of the recursion could be equal to the number of nodes N in the tree. Each recursive call adds a new frame to the stack to store function parameters and local variables. Therefore, the auxiliary space used by the recursion stack can grow linearly with the number of nodes in the tree, leading to O(N) space complexity.

Edge Cases

Case	How to Handle
nodes is 0 or less	Return 0 immediately, as an empty tree has no nodes.
parent is null or empty	If nodes is greater than 0, consider this an invalid input and return 0, otherwise return 0 which is already handled in the first edge case.
value is null or empty	If nodes is greater than 0, consider this an invalid input and return 0, otherwise return 0 which is already handled in the first edge case.
Length of parent and value arrays are not equal to nodes	Consider this an invalid input, returning 0.
A node is its own parent (cycle in graph)	The algorithm should detect cycles and avoid infinite loops, possibly by marking visited nodes.
Integer overflow in sum calculation	Use a larger integer type (e.g., long) to store subtree sums to prevent overflow.
All node values are zero	The algorithm correctly removes all nodes if all subtrees have a sum of zero, leaving zero nodes.
The tree is a single path (highly skewed distribution)	The recursive or iterative tree traversal should handle long paths without stack overflow (iterative preferred) or excessive memory usage.

nodes is 0 or less

How to Handle:

Return 0 immediately, as an empty tree has no nodes.

parent is null or empty

How to Handle:

If nodes is greater than 0, consider this an invalid input and return 0, otherwise return 0 which is already handled in the first edge case.

value is null or empty

How to Handle:

If nodes is greater than 0, consider this an invalid input and return 0, otherwise return 0 which is already handled in the first edge case.

Length of parent and value arrays are not equal to nodes

How to Handle:

Consider this an invalid input, returning 0.

A node is its own parent (cycle in graph)

How to Handle:

The algorithm should detect cycles and avoid infinite loops, possibly by marking visited nodes.

Integer overflow in sum calculation

How to Handle:

Use a larger integer type (e.g., long) to store subtree sums to prevent overflow.

All node values are zero

How to Handle:

The algorithm correctly removes all nodes if all subtrees have a sum of zero, leaving zero nodes.

The tree is a single path (highly skewed distribution)

How to Handle:

The recursive or iterative tree traversal should handle long paths without stack overflow (iterative preferred) or excessive memory usage.