Binary Tree Tilt

To calculate the sum of every tree node’s tilt, we can use a recursive function that traverses the tree in a bottom-up manner. Here’s the code followed by an explanation:

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class Solution:
    def findTilt(self, root: Optional[TreeNode]) -> int:
        self.total_tilt = 0

        def subtree_sum(node):
            if not node:
                return 0
            left_sum = subtree_sum(node.left)
            right_sum = subtree_sum(node.right)
            tilt = abs(left_sum - right_sum)
            self.total_tilt += tilt
            return node.val + left_sum + right_sum

        subtree_sum(root)
        return self.total_tilt

Explanation

  1. Initialize Tilt Variable: self.total_tilt is a class variable to keep track of the total tilt of the tree.

  2. Define Recursive Function: The function subtree_sum calculates the sum of the values of the nodes in the subtree rooted at a given node and at the same time computes the tilt for that node. The base case is when the node is None, in which case the sum is 0.

  3. Calculate Left and Right Sum: For a given node, the function recursively calculates the sum of the values of the nodes in the left and right subtrees (left_sum and right_sum).

  4. Compute Tilt: The tilt of a node is the absolute difference between left_sum and right_sum. The tilt is added to the total tilt of the tree.

  5. Return Subtree Sum: The function returns the sum of the values of the nodes in the subtree rooted at the given node, including the value of the given node itself.

  6. Invoke Recursive Function: The function subtree_sum is invoked on the root of the tree.

  7. Return Total Tilt: Finally, the total tilt of the tree is returned.

Insights

  • Recursive Traversal: The tree is traversed in a bottom-up manner, ensuring that the tilt of each node is computed only after the tilts of its children have been computed.
  • Time Complexity: This solution has a time complexity of (O(n)), where (n) is the number of nodes in the tree, as each node is visited once.
  • Space Complexity: The space complexity is (O(h)), where (h) is the height of the tree, due to the recursive call stack. In the worst case, this is (O(n)) for a skewed tree and (O(\log n)) for a balanced tree.
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# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val = 0, left = nil, right = nil)
#         @val = val
#         @left = left
#         @right = right
#     end
# end
# @param {TreeNode} root
# @return {Integer}

  
  Define the Terms
  
  1. sum(left subtree node values) - sum(right subtree node values)
  2. If no child => sum = 0
  3. tilt = abs(#1)
  
  Thinking Recursively
  
  1. What is the unit of work to be done in each recursive call?
        - Compute the tilt of the current node
  2. How do we reduce the input size in the recursive call?
        - Binary tree is recursive, split the left and right subtrees separately
  3. Return the sum of every node's tilt
        - Need to sum all the values
        - This needs to survive between different stack frames
  
  Problem Decomposition
  
  1. Compute tilt for each node
  2. Sum all the tilts
  
  
  Gotchas
  
  1. leaf node and base case are different

```ruby
def tilt_wrapper(root)
#     base case
    if root.nil?
        return 0
    end
#     leaf node
    if root.left.nil? && root.right.nil?
        return root.val
    end

    left = tilt_wrapper(root.left)
    right = tilt_wrapper(root.right)

    @sum += (left - right).abs

    return left + right + root.val
end

def find_tilt(root)
    @sum = 0
    tilt_wrapper(root)
    @sum
end

Identifying Problem Isomorphism

“Binary Tree Tilt” can be approximately mapped to “Diameter of Binary Tree”.

Both involve calculating a value based on the properties of a binary tree’s nodes and their relationships. In “Binary Tree Tilt”, the goal is to find the tilt of the entire tree, defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values for each node. In “Diameter of Binary Tree”, you’re asked to compute the length of the longest path between any two nodes, which is essentially the maximum distance between two nodes.

Both require the use of depth-first search (DFS) to traverse the tree and some form of postorder traversal to accumulate values up the tree. The primary difference lies in what values are being accumulated and how they are processed. In “Binary Tree Tilt”, you’re looking at the sums of subtrees, whereas in “Diameter of Binary Tree”, you’re considering path lengths.

“Diameter of Binary Tree” is simpler as it doesn’t require handling of the absolute difference, but it does require understanding the concept of the longest path in a binary tree, which can be challenging for beginners. However, the complexity difference is not significant.

Before tackling the “Binary Tree Tilt” problem (LeetCode 563), understand the basic concepts of Binary Trees and the idea of recursion in depth-first traversal. Here are some problems that can help build up these fundamentals:

  1. “Maximum Depth of Binary Tree” (LeetCode 104): Understand the basics of depth-first traversal in a binary tree.

  2. “Validate Binary Search Tree” (LeetCode 98): Get familiar with the properties of Binary Search Trees (BSTs).

  3. “Symmetric Tree” (LeetCode 101): Understand the concept of tree symmetry and how to identify it.

  4. “Path Sum” (LeetCode 112): Understand how to trace a path in a tree with a given total sum.

  5. “Binary Tree Level Order Traversal” (LeetCode 102): Understand breadth-first traversal in a binary tree.

  6. “Invert Binary Tree” (LeetCode 226): Practice manipulating the tree structure.

  7. “Convert Sorted Array to Binary Search Tree” (LeetCode 108): Understand how a BST can be built from a sorted array.

  8. “Binary Tree Paths” (LeetCode 257): Learn how to trace all paths from the root to leaves in a binary tree.

  9. “Subtree of Another Tree” (LeetCode 572): Understand the concept of a subtree and how to check for it.

  10. “Diameter of Binary Tree” (LeetCode 543): Learn how to calculate the diameter of a binary tree which can help understand the concept of tilt.

These cover Binary Trees and their common manipulations, which will help you tackle the “Binary Tree Tilt” problem effectively.

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class Solution:
    def findTilt(self, root: TreeNode) -> int:
        def helper(root):
            if not root: return 0
            lv, rv = helper(root.left), helper(root.right)
            self.ans += abs(lv - rv)
            return root.val + lv + rv

        self.ans = 0
        helper(root)
        return self.ans

Problem Classification

This problem belongs to the “Binary Trees”. The task is centered around traversing and processing the nodes of a binary tree, which is a fundamental concept in this domain.

‘What’ Components:

  1. A binary tree: This data structure is central to the problem. Each node in the tree has at most two children (left and right), and each node holds a value.

  2. The tilt of a tree node: This is a derived property of each node in the tree, defined as the absolute difference between the sum of all left subtree node values and all right subtree node values.

  3. The sum of all tilts: This is the ultimate desired outcome - we need to compute the tilt for each node and sum all these tilts.

Given the task requires navigating and manipulating a binary tree, the problem can be classified as one of “Tree Traversal”. Specifically, it’s a variant of a “Depth-First Search” (DFS) problem where we are visiting every node to calculate a derived value (the tilt), and accumulating this value across all nodes. The depth-first traversal is implied by the need to calculate the sum of the values of all children of a node (which might themselves be roots of sub-trees), suggesting a post-order traversal might be useful (since we want to process a node only after its children).

Clarification Questions

What are the clarification questions we can ask about this problem?

Language Agnostic Coding Drills

  1. Dissect the Code:

This piece of code employs several different coding concepts:

a. Recursion: The helper function is a recursive function, a common approach to tree-based problems. The recursion in this case is a post-order traversal of the tree (left subtree -> right subtree -> root node).

b. Conditional Statements: The code uses a conditional statement to check if a node is null, a key step in any binary tree traversal to determine when the recursion should stop.

c. Arithmetic Operations: The code uses basic arithmetic operations to calculate the tilt of a node and sum of all tilts.

d. Variable Initialization and Assignment: The code uses variables to track the tilt sum (self.ans), the sums of the left and right subtrees (lv, rv), and to return the sum of a node’s value and its subtrees.

e. Use of self in classes: The use of “self” is crucial for maintaining the total tilt value across all recursive calls.

  1. Coding Concepts in Order of Increasing Difficulty:

a. Variable Initialization and Assignment: This is a fundamental coding concept, common to all programming languages.

b. Arithmetic Operations: This is another basic coding concept. In this case, the arithmetic is straightforward (addition and subtraction).

c. Conditional Statements: This concept is slightly more advanced, as it requires understanding of Boolean expressions and control flow.

d. Use of self in classes: This is an object-oriented programming concept that is medium difficulty. It requires understanding how object instances maintain their state.

e. Recursion: This is one of the more complex concepts, particularly in this case where it’s used for tree traversal. Understanding recursion requires comprehending how function calls are placed on the call stack and popped off as they complete, and how return values are passed up the call stack.

  1. Problem-solving approach:

The problem-solving strategy here is rooted in a post-order depth-first search, where each node’s value and its children’s values are summed up and the absolute difference between the left and right children’s sums (the tilt) is added to a total. The steps to solve this problem are:

a. First, set up a variable to hold the total tilt of all nodes.

b. Start a post-order depth-first search on the binary tree.

c. For each node, recursively find the sum of its left and right subtrees.

d. Calculate the tilt of the current node (the absolute difference between the sum of the left subtree and the sum of the right subtree).

e. Add the node’s tilt to the total tilt.

f. Return the sum of the node’s value and the sums of its left and right subtrees.

By repeating this process for each node in the tree, starting from the root, the function calculates the total tilt of all nodes in the binary tree.

Targeted Drills in Python

  1. Python-based Coding Drills:

a. Variable Initialization and Assignment:

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# Initialize a variable
a = 0
# Assign a new value to the variable
a = 5

b. Arithmetic Operations:

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# Addition
sum_val = 3 + 2  # sum_val now holds the value 5

# Subtraction
diff_val = 3 - 2  # diff_val now holds the value 1

# Absolute value
abs_val = abs(-5)  # abs_val now holds the value 5

c. Conditional Statements:

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# If-else condition
a = 5
if a > 0:
    print("a is positive")
else:
    print("a is non-positive")

d. Use of self in classes:

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class MyClass:
    def __init__(self):
        self.variable = 0

    def update_variable(self, new_val):
        self.variable = new_val

e. Recursion:

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def factorial(n):
    if n == 0:  # Base case
        return 1
    else:
        return n * factorial(n-1)  # Recursive call
  1. Problem-specific drills:

One of the main problem-specific concepts in this problem is the traversal of a binary tree. This can be done using recursion. Here is a simple drill to illustrate a tree traversal (In-Order Traversal):

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class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

def inorder_traversal(root):
    if root:
        inorder_traversal(root.left)
        print(root.val)
        inorder_traversal(root.right)
  1. Integrating the Drills:

All the concepts illustrated above can be integrated to solve the initial problem in the following order:

a. Begin by initializing a class variable (self.ans) to hold the total tilt.

b. Define a helper function (helper) to perform a post-order traversal of the tree using recursion.

c. Inside the helper function, use conditional statements to handle the case where the node is None.

d. Call the helper function recursively on the left and right children of the current node to calculate the sum of the left and right subtrees. Store these sums in variables.

e. Use arithmetic operations to calculate the tilt of the current node and add it to the total tilt.

f. Finally, return the sum of the node’s value and the sums of its left and right subtrees.

By calling the helper function on the root of the tree, you initiate the process of calculating the total tilt of all nodes in the tree.