Introduction

In this blog post, we will explore how to solve the LeetCode problem "951" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees. A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations. Given the roots of two binary trees root1 and root2, return true if the two trees are flip equivalent or 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 Constraints: The number of nodes in each tree is in the range [0, 100]. Each tree will have unique node values in the range [0, 99].

Explanation

Here's a solution to determine if two binary trees are flip equivalent:

• Core Idea: Recursively check if two trees are flip equivalent. At each node, check if the values are equal. Then, there are two possibilities: either the left and right subtrees match in the same order, or they are flipped. Recursively check both possibilities.
• Base Cases: Empty trees are equivalent. If one tree is empty and the other isn't, they are not equivalent. If node values are different, they are not equivalent.

• Recursive Step: Recursively check if the left of tree1 is equivalent to the left of tree2 AND the right of tree1 is equivalent to the right of tree2, OR the left of tree1 is equivalent to the right of tree2 AND the right of tree1 is equivalent to the left of tree2.

• Complexity: O(N), where N is the number of nodes in the smaller tree. O(H) space for the recursion stack, where H is the height of the smaller tree. In the worst case, H can be N for skewed trees.

Code

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

def flipEquiv(root1, root2):
    if not root1 and not root2:
        return True
    if not root1 or not root2 or root1.val != root2.val:
        return False

    return (flipEquiv(root1.left, root2.left) and flipEquiv(root1.right, root2.right)) or \
           (flipEquiv(root1.left, root2.right) and flipEquiv(root1.right, root2.left))