Make Costs of Paths Equal in a Binary Tree

This problem requires us to determine the minimum increments needed to make the cost of paths from the root to each leaf node equal in a perfect binary tree.

For each node in the tree, there are two paths leading to its two children. We should equalize the total cost of these two paths first. Once the costs of two paths originating from a node become equal, the node’s path sum becomes equal to the maximum path sum of its children.

After this, we recursively adjust its parent’s path sums until we reach the root. Finally, all the paths leading from the root to leaf nodes will have equal costs.

Steps to solve the problem:

  1. Use a recursive function dfs to traverse the tree.
  2. For each node, get the cost of paths for its left and right children.
  3. Calculate the difference between these two paths.
  4. The absolute difference is the number of increments needed to equalize the paths from the current node. Add this value to the answer.
  5. Return the maximum path sum of its children plus the current node’s cost.

Let’s code this solution:

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class Solution:
    def minIncrements(self, n: int, cost: List[int]) -> int:
        # Recursive function to traverse the tree and get the total path sum for each node
        def dfs(node: int) -> int:
            if node > n:  # base case: if the node is out of bounds
                return 0

            # Get the total path sum of the left and right child
            left_sum = dfs(2 * node) 
            right_sum = dfs(2 * node + 1)

            # Add the absolute difference between the two path sums to the answer
            self.res += abs(left_sum - right_sum)

            # Return the maximum path sum of the two children plus the current node's cost
            return max(left_sum, right_sum) + cost[node - 1]

        self.res = 0  # Initialize the result
        dfs(1)  # Start the DFS from the root
        return self.res

The dfs function returns the total path sum for each node, and the difference between the two children’s path sums is added to the result.

10 Prerequisite LeetCode Problems

“Make Costs of Paths Equal in a Binary Tree” involves adjusting the weights of nodes in a binary tree so that the total cost (sum of node values) of every path from root to leaf is the same.

An approximate isomorphic problem is “Equal Tree Partition”. Here, the task is to decide whether you can partition the tree into two trees which have equal sum of values even if you remove exactly one edge from the tree.

Both problems involve working with a binary tree and the concept of maintaining balance in some form, either in terms of the path cost or the sum of values. In the first, we are modifying the tree to achieve balance, and in the second, we’re determining whether the tree can be partitioned into balanced parts.

These problems are not exactly identical. “Make Costs of Paths Equal in a Binary Tree” deals with modifying the tree, while “Equal Tree Partition” is more about finding a particular property in the tree.

“Make Costs of Paths Equal in a Binary Tree” is more complex due to the nature of modifications it involves. It requires a deep understanding of tree traversals and handling the distribution of weights. “Equal Tree Partition”, being the simpler problem, can provide an understanding of how to approach the more complex one, especially in terms of tree traversal and summing up the node values.

10 Prerequisite LeetCode Problems

For “2673. Make Costs of Paths Equal in a Binary Tree”, the following are a good preparation:

  1. “108. Convert Sorted Array to Binary Search Tree” - Understanding basic binary tree construction which is the foundational step of solving this problem.

  2. “101. Symmetric Tree” - Practicing traversing a binary tree and comparing nodes in a binary tree.

  3. “112. Path Sum” - Understanding how to calculate path sums in a binary tree, which is a key component of this problem.

  4. “226. Invert Binary Tree” - Practicing manipulating the structure of a binary tree, which could be useful in the main problem.

  5. “104. Maximum Depth of Binary Tree” - Understanding depth of a binary tree, this understanding would help in solving this problem as we are dealing with perfect binary trees.

  6. “102. Binary Tree Level Order Traversal” - Getting comfortable with level order traversal, which could be useful for visiting each node in the main problem.

  7. “110. Balanced Binary Tree” - Another problem related to the structure of binary trees, which could be useful in the main problem.

  8. “563. Binary Tree Tilt” - This problem involves calculating a value based on the difference between subtrees, which is similar to how the main problem involves calculating a value based on the differences between paths.

  9. “671. Second Minimum Node In a Binary Tree” - Helps in understanding concepts of node manipulation in binary tree.

  10. “129. Sum Root to Leaf Numbers” - This problem practices summing values along paths in a binary tree, a key operation in the main problem.

These cover the key concepts of binary tree traversal, manipulation, and properties of perfect binary trees, which will be helpful in tackling the main problem.

You are given an integer n representing the number of nodes in a perfect binary tree consisting of nodes numbered from 1 to n. The root of the tree is node 1 and each node i in the tree has two children where the left child is the node 2 * i and the right child is 2 * i + 1.

Each node in the tree also has a cost represented by a given 0-indexed integer array cost of size n where cost[i] is the cost of node i + 1. You are allowed to increment the cost of any node by 1 any number of times.

Return the minimum number of increments you need to make the cost of paths from the root to each leaf node equal.

Note:

A perfect binary tree is a tree where each node, except the leaf nodes, has exactly 2 children. The cost of a path is the sum of costs of nodes in the path.

Example 1:

Input: n = 7, cost = [1,5,2,2,3,3,1] Output: 6 Explanation: We can do the following increments:

  • Increase the cost of node 4 one time.
  • Increase the cost of node 3 three times.
  • Increase the cost of node 7 two times. Each path from the root to a leaf will have a total cost of 9. The total increments we did is 1 + 3 + 2 = 6. It can be shown that this is the minimum answer we can achieve.

Example 2:

Input: n = 3, cost = [5,3,3] Output: 0 Explanation: The two paths already have equal total costs, so no increments are needed.

Constraints:

3 <= n <= 105 n + 1 is a power of 2 cost.length == n 1 <= cost[i] <= 104

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def minIncrements(self, n, cost):
    self.res = 0
    def dfs(i):
        if i >= len(cost): return 0
        a, b = dfs(2 * i + 1), dfs(2 * i + 2)
        self.res += abs(a - b)
        return cost[i] + max(a, b)
    dfs(0)
    return self.res

Problem Classification

The problem statement belongs to the domain of computer science, particularly focusing on data structures and algorithms. More specifically, it is related to binary trees and path cost computation.

The ‘What’ components of the problem are:

  1. Perfect Binary Tree: This is a type of binary tree in which all interior nodes have two children and all leaves have the same depth or same level. In the context of this problem, the tree has ’n’ nodes with specific numbering as per the problem statement.

  2. Node Cost: Each node in the tree has an associated cost, which is provided as an array of integers.

  3. Incrementing Cost: The cost of any node in the tree can be increased by 1 as many times as needed.

  4. Path Cost: The cost of a path from the root to a leaf is defined as the sum of the costs of all nodes in the path.

  5. Minimum Increments: The objective is to find the minimum number of increments required to make the cost of paths from the root to each leaf node equal.

Based on the identified components, the problem can be classified as a decision-making problem that involves optimization (minimizing the increments) under certain conditions (making path costs equal). The problem requires a good understanding of binary trees and tree traversal algorithms. A suitable algorithm needs to be developed that takes into account the cost at each node, the ability to increment node costs, and the need to equalize the path costs.

Language Agnostic Coding Drills

  1. Dissection of the Code into Distinct Concepts:

The provided Python code implements a solution to the problem using recursion, specifically a depth-first search (DFS) technique. The following distinct coding concepts are found in the code:

  • Function Definition: Defining a Python function with parameters.

  • Recursion: Using recursion to solve a problem, in this case using depth-first search.

  • Conditional Statements: Using an if statement to control the flow of the program.

  • Arithmetic Operations: Simple arithmetic operations like addition, multiplication and taking absolute value.

  • Returning Values: Returning a value from a function.

  1. Concepts in Order of Increasing Difficulty:

    • Function Definition: This is a basic concept in most programming languages. A user-defined function is created with the def keyword in Python.

    • Arithmetic Operations: These are also basic operations in programming and include addition, subtraction, multiplication, etc.

    • Returning Values: A function can return a value as its result after processing the function body.

    • Conditional Statements: Conditional statements control the flow of execution based on certain conditions. In Python, the keywords if, elif, and else are used.

    • Recursion: Recursion can be a complex concept for beginners. It involves a function calling itself as part of the solution. It is often used in problems involving tree or graph traversal, as in this case.

  2. Problem-Solving Approach:

The problem-solving approach uses depth-first search to traverse the perfect binary tree represented in an array. The dfs function is used to perform the recursive traversal.

The base case of the recursion is when i is greater than or equal to the length of the cost array, at which point the recursion returns 0. For each node, the function calculates the costs for the left child a and right child b by recursively calling dfs.

The absolute difference between the costs of the left and right subtrees is added to the global res variable which keeps track of the total increments. For each node, the function returns the cost of the node plus the maximum of the costs of its left and right subtrees. This ensures that all paths from the root to the leaves have the same total cost. The main function then returns res as the minimum number of increments needed.

Targeted Drills in Python

  1. Python Code for Identified Concepts:

    • Function Definition:

      A Python function is defined using the def keyword. Here is an example:

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      def greet():
    print("Hello, world!")

    • Arithmetic Operations:

      Arithmetic operations in Python include addition, subtraction, multiplication, etc. Here is an example:

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      a = 10
b = 20
print(a + b)  # Output: 30

    • Returning Values:

      A function can return a value as its result after processing the function body. Here is an example:

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      def add(a, b):
    return a + b

    • Conditional Statements:

      Conditional statements in Python are used to control the flow of execution. Here is an example:

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      a = 10
if a > 0:
    print("a is positive.")

    • Recursion:

      Recursion in Python is a process where a function calls itself as its subroutine. Here is an example of a recursive function to find the factorial of a number:

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      def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n - 1)

  2. Problem-Specific Concepts:

    • Depth-First Search (DFS):

      DFS is a tree/graph traversal method which is used extensively in the original problem to traverse the binary tree. DFS visits the child nodes before visiting the sibling nodes; that is, it traverses the depth of any particular path before exploring its breadth. Here is a simple Python code snippet that encapsulates this concept:

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      def dfs(node):
    if node is None:
        return
    print(node.value)
    dfs(node.left)
    dfs(node.right)

    • Array Representation of Binary Tree:

      The problem uses an array representation of a binary tree where for a node at index i, its left child is at index 2*i+1 and right child is at index 2*i+2. This concept is crucial for understanding the given problem.

  3. Assembling the Pieces:

    Each of the above pieces are integral to solving the problem. Start by understanding how binary trees can be represented using arrays. This will help in visualising the problem. Then learn about Depth-First Search as it is used to traverse the tree in the problem. Understand the other basic concepts like arithmetic operations, conditional statements, function definitions and returning values from a function.

    In the problem, the function dfs is defined inside the main function. It traverses the tree using DFS. For each node, it recursively calculates the costs of the left and right subtrees, adds the absolute difference to the global result, and then returns the cost of the node plus the maximum cost of its subtrees. This process continues until all nodes have been visited and the final result is returned.

    Hence, the individual drills not only help in understanding the original problem better but also facilitate in building up to the final solution. Each concept is applied at the right place and in the right order to solve the problem.