How do you solve Binary Tree Tilt on LeetCode?

Binary Tree Tilt (LeetCode #563) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: Using a single traversal reduces redundant calculations.

What is the brute force solution for Binary Tree Tilt?

The brute force approach for Binary Tree Tilt is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach calculates the tilt for each node by separately summing the values of its left and right subtrees. This method is straightforward but inefficient as it repeatedly traverses the same nodes. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Binary Tree Tilt?

Binary Tree Tilt uses the following concepts: Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Tree Traversal.

What are the interview tips for Binary Tree Tilt?

Explain your thought process clearly while coding. Consider edge cases, like empty trees or single-node trees. Discuss time and space complexities after presenting your solution.

What are common mistakes when solving Binary Tree Tilt?

Not considering null children when calculating sums. Forgetting to use absolute values when calculating tilt.

#563

Binary Tree Tilt

Easy

TreeDepth-First SearchBinary TreeDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(h)

💡

Intuition

Time O(n)Space O(h)

The optimal solution uses a single traversal of the tree to calculate both the tilt and the sum of the subtree values. This avoids redundant calculations and improves efficiency.

⚙️

Algorithm

4 steps

1Step 1: Traverse the tree using a depth-first search (DFS).
2Step 2: For each node, calculate the sum of its left and right subtrees in one pass.
3Step 3: Calculate the tilt and add it to a global sum during the traversal.
4Step 4: Return the total tilt after the traversal is complete.

solution.py20 lines

1class TreeNode:
2    def __init__(self, val=0, left=None, right=None):
3        self.val = val
4        self.left = left
5        self.right = right
6
7class Solution:
8    def findTilt(self, root: TreeNode) -> int:
9        def dfs(node):
10            if not node:
11                return 0
12            left_sum = dfs(node.left)
13            right_sum = dfs(node.right)
14            tilt = abs(left_sum - right_sum)
15            self.total_tilt += tilt
16            return left_sum + right_sum + node.val
17
18        self.total_tilt = 0
19        dfs(root)
20        return self.total_tilt

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1Using a single traversal reduces redundant calculations.
2Understanding tree structure helps in visualizing subtree sums.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.