How do you solve Flip Equivalent Binary Trees on LeetCode?

Flip Equivalent Binary Trees (LeetCode #951) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Flipping a tree is similar to rearranging furniture; the same pieces can look different based on how they are arranged.

What is the brute force solution for Flip Equivalent Binary Trees?

The brute force approach for Flip Equivalent Binary Trees is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible configuration of the trees by recursively comparing nodes and their children. If we encounter a mismatch, we can try flipping the children and check again. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Flip Equivalent Binary Trees?

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

What are the interview tips for Flip Equivalent Binary Trees?

Always clarify the problem statement and edge cases before diving into coding. Think about the recursive nature of trees; visualize the tree structure as you write your solution. Discuss your thought process with the interviewer; it shows your understanding and can lead to hints.

What are common mistakes when solving Flip Equivalent Binary Trees?

Not considering the case where one tree is null and the other is not. Failing to check both configurations (flipped and non-flipped) properly.

#951

Flip Equivalent Binary Trees

Medium

TreeDepth-First SearchBinary TreeRecursionTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses recursion but avoids unnecessary checks by directly comparing the nodes and their children in a single pass. This reduces redundant calculations and improves efficiency.

⚙️

Algorithm

4 steps

1Step 1: If both nodes are null, return true.
2Step 2: If one node is null and the other is not, return false.
3Step 3: If the values of the nodes are different, return false.
4Step 4: Recursively check both configurations without flipping and with flipping, but only once for each configuration.

solution.py8 lines

1def flipEquiv(root1, root2):
2    if not root1 and not root2:
3        return True
4    if not root1 or not root2:
5        return False
6    if root1.val != root2.val:
7        return False
8    return (flipEquiv(root1.left, root2.left) and flipEquiv(root1.right, root2.right)) or (flipEquiv(root1.left, root2.right) and flipEquiv(root1.right, root2.left))

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once, and the space complexity is O(n) due to the recursion stack in the worst case.

1Flipping a tree is similar to rearranging furniture; the same pieces can look different based on how they are arranged.
2The problem can be approached recursively, leveraging the properties of binary trees.

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