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Calculate Tilt

easy
DESCRIPTION (inspired by Leetcode.com)

Given the root node of a binary tree, write a recursive function to return the sum of all node tilts in the tree.

The tilt of a node is calculated as the absolute difference between the sum of all values in its left subtree and the sum of all values in its right subtree. For nodes that are missing a left child, right child, or both, treat the missing subtree as having a sum of 0.

For example, a leaf node has a tilt of 0 because both its left and right subtrees are empty (sum = 0), so the absolute difference is |0 - 0| = 0.

Example 1:

513

Input:

[5, 1, 3]

Output:

2

The leaf nodes 1, 3 have tilts of 0 (their left and right subtrees are empty)

The root node 5 has a tilt of |1 - 3| = 2

Example 2:

4213769

Input:

[4, 2, 7, 1, 3, 6, 9]

Output:

21

The leaf nodes 1, 3, 6, 9 have tilts of 0 (their left and right subtrees are empty)

Node 2 has a tilt of |1 - 3| = 2 Node 7 has a tilt of |6 - 9| = 3 Node 4 has a tilt of |6 - 22] = 16

2 + 3 + 16 = 21

Explanation

We can solve this problem using the framework established in the Overview section.
Return Values
If I'm at a node in the tree, what values do I need to calculate the tilt of the subtree rooted at that node?
  1. The sum of the values in the left subtree.
  2. The sum of the values in the right subtree.
This tells me that each recursive call should return the sum of the values in the subtree rooted at the current node. So in the body of the recursive function, I'll make recursive calls to the current node's left and right children, and then add the value of the current node to return the sum of the current subtree.
Before returning the sum of the current subtree, I'll also need to calculate the tilt of the current subtree. I can calculate the tilt of the current subtree as abs(left - right), and add that a global variable tilt that stores the total tilt of the tree. (see the Global Variables section for more details on why this is necessary).
Solution
# get the sum of current node's left and right subtrees
left = dfs(node.left)
right = dfs(node.right)
# calculate tilt of current subtree and add it to the global tilt variable
tilt += abs(left - right)
# return the sum of the current subtree
return left + right + node.val
Base Case
The tilt of an empty tree is 0.
Global Variables Note that the problem is asking for the total tilt of the tree, but each recursive call returns the sum of the nodes in the subtree rooted at a given node. Since these two values are different, we can use a global variable tilt to store the total tilt of the tree. This allows us to avoid using multiple return values in our recursive function, or having to pass the total tilt value from the parent to the child.
Helper Function
We don't need to pass values from the parent to the child to solve this problem. However, since we are using a global variable to store the total tilt of the tree, we should define tilt inside the body of our main function, and then introduce a helper function to perform the recursive calls. This way, no code outside of the main function will have access to the tilt variable.
Inside the main function, we can initiate the call to the helper function, passing in the root of the tree as an argument. We return the value of tilt after the recursive helper function has finished executing.

Solution

Solution
class Solution:
    def findTilt(self, root: TreeNode) -> int:
        tilt = 0
       # define a helper function to perform the recursive calls 
       # this ensures that the tilt variable is not accessible outside of the main function
        def dfs(node):
            nonlocal tilt
            # base case
            if not node:
                return 0
            # get the sum of the current node's left and right subtrees
            left = dfs(node.left)
            right = dfs(node.right)
            # calculate tilt of current subtree, and add it to the global tilt variable
            tilt += abs(left - right)
            # return the sum of the current subtree
            return left + right + node.val
       # initiate the call to the helper function 
        dfs(root)
        return tilt
Test Your Knowledge

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Complexity Analysis

Time Complexity: O(N) where N is the number of nodes in the tree. We visit each node exactly once, and at each node, we perform constant time work.

Space Complexity: O(N) where N is the number of nodes. A total of N call frames have to be allocated, one for each node in the tree.

Mark as read
Next: Diameter of a Binary Tree

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