Welcome back, coders! 👨‍💻👩‍💻 Today, we’re going to dive deep into a fascinating problem that involves binary trees, recursion, and some nifty tree manipulations – specifically, flip operations! 🌲🔀🌲 In this post, we’ll learn how to check if two binary trees are flip equivalent, meaning that by flipping some nodes’ children, we can make one tree look like the other. 🌀

Get ready to explore this problem 😄

Problem Statement 📝

Given two binary trees root1 and root2, we want to determine if they are flip equivalent. A binary tree is flip equivalent to another if we can make them identical by flipping (swapping the left and right child subtrees) at some nodes. A flip operation can be done any number of times, and it can be performed at any node in the tree.

Our task is to write a function that returns True if the two trees are flip equivalent, and False otherwise.

Example 1:

Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], 
       root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.

Example 2:

Input: root1 = [], root2 = []
Output: true

Example 3:

Input: root1 = [], root2 = [1]
Output: false

🌟 Basic Solution 🌟

The problem is essentially about comparing the structure of two binary trees while allowing for flips. This is a natural candidate for recursion because each node comparison depends on the outcome of its child nodes. Here’s how we can break it down:

1. Base Case 1: If both root1 and root2 are None, we return True. This means we’ve reached the end of both trees, so they are trivially equivalent.

2. Base Case 2: If one of the trees is None and the other is not, they can’t be equivalent, so we return False.

3. Value Comparison: If the values of the nodes don’t match, return False.

4. Recursive Check on Children: We need to check if the subtrees of both trees are either:

   • Equivalent without flipping, i.e., the left child of root1 matches the left child of root2 and the right child of root1 matches the right child of root2.
   • Equivalent after flipping, i.e., the left child of root1 matches the right child of root2, and the right child of root1 matches the left child of root2.

If either condition holds, then the trees are flip equivalent!

Python Code 🐍

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def flip_equiv(root1: TreeNode, root2: TreeNode) -> bool:
    # Base case: both are None
    if not root1 and not root2:
        return True
    # One is None, but the other isn't
    if not root1 or not root2:
        return False
    # Values don't match
    if root1.val != root2.val:
        return False

    # Check if children are equivalent directly or after flipping
    return (flip_equiv(root1.left, root2.left) and flip_equiv(root1.right, root2.right)) or \
           (flip_equiv(root1.left, root2.right) and flip_equiv(root1.right, root2.left))

Explanation:

• The function flip_equiv takes two tree roots (root1 and root2) as input.
• It checks the base cases first (both None, or one None while the other is not).
• If the current nodes have the same value, it recursively checks the subtrees:
  • Either the subtrees are directly equivalent, or they become equivalent after a flip.

⏱️ Time Complexity of Basic Solution ⏱️

The time complexity of this solution is O(min(N1, N2)), where N1 and N2 are the number of nodes in the two trees.

• In the worst case, we will need to traverse both trees entirely, comparing each node with its counterpart in the other tree.
• Since we only need to compare corresponding nodes and their children once, the solution is efficient for most cases.

🌟 Optimized Solution 🌟

Since the basic solution already efficiently solves the problem using recursion, no further optimization is necessary. The recursion explores all possible ways of flipping the child nodes and ensures that each node is visited exactly once.

The algorithm has an early exit mechanism with the base cases that allow us to stop further computation as soon as we know that the trees are not flip equivalent. This makes the recursive approach both optimal and easy to understand.

Since the time complexity remains O(min(N1, N2)), we can conclude that the recursive solution is already optimal for this problem.

🧑‍🏫 Example Walkthrough with Higher Numbers 🧑‍🏫

Let’s walk through an example to understand the solution better. We’ll use a larger example to illustrate how the recursion works in more depth.

Example:

root1 = [10, 20, 30, 40, 50, null, null, 70, 80, 90, 100]
root2 = [10, 30, 20, null, null, 40, 50, 100, 90, 80, 70]

Step-by-Step Walkthrough:

1. Compare root nodes (10):
   • Both trees have the root node with value 10. Since the values are equal, we proceed to compare the children.
2. Compare left and right children of 10:
   • In root1, the left child is 20 and the right child is 30.
   • In root2, the left child is 30 and the right child is 20.
   • The children don’t match directly, so we try flipping: we compare root1’s left child (20) with root2’s right child (20), and root1’s right child (30) with root2’s left child (30). Both comparisons hold, so we proceed!
3. Compare subtrees rooted at 20:
   • In root1, node 20 has left child 40 and right child 50.
   • In root2, node 20 has left child 50 and right child 40.
   • Again, these don’t match directly, but after flipping, we see that:
     • root1.left.left (node 40) matches root2.left.right (node 40), and
     • root1.left.right (node 50) matches root2.left.left (node 50).
   • Both subtrees are equivalent after flipping, so we move on.
4. Compare subtrees rooted at 50:
   • In root1, node 50 has left child 90 and right child 100.
   • In root2, node 50 has left child 100 and right child 90.
   • These don’t match directly, but after flipping, we see:
     • root1.left.right.left (node 90) matches root2.left.left.right (node 90), and
     • root1.left.right.right (node 100) matches root2.left.left.left (node 100).
   • The subtrees are equivalent after flipping.

At each level of the tree, the recursive approach checks if a flip operation is needed, ensuring that the trees are compared properly.

🏁 Conclusion 🏁

The Flip Equivalent Binary Trees problem demonstrates the power of recursion in solving problems involving hierarchical data structures like trees. By using recursion, we can systematically explore all possible flip operations and determine whether two trees are equivalent.

We learned that:

• The base case checks for None values and unequal node values are essential to stop the recursion early.
• The recursive approach efficiently handles the comparison by exploring both direct and flipped possibilities.
• The time complexity of the solution is O(min(N1, N2)), making it optimal.

In the end, understanding how to manipulate binary trees and apply recursive thinking will make you a stronger problem solver in many areas of computer science! 💡🌳