Flip Equivalent Binary Trees - Python Solution Leetcode

 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 equivelent 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

Example 4:

Input: root1 = [0,null,1], root2 = []
Output: false

Example 5:

Input: root1 = [0,null,1], root2 = [0,1]
Output: true

Solution #1

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flipEquiv(self, root1: TreeNode, root2: TreeNode) -> bool:
        if root1 is root2:
            return True
        if not root1 or not root2 or root1.val != root2.val:
            return False
        return self.flipEquiv(root1.left,root2.left) and self.flipEquiv(root1.right,root2.right) or self.flipEquiv(root1.left,root2.right) and self.flipEquiv(root1.right,root2.left)
        

Aproach #2

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flipEquiv(self, root1, root2):
        def dfs(node):
            if node:
                yield node.val
                L = node.left.val if node.left else -1
                R = node.right.val if node.right else -1
                if L < R:
                    yield from dfs(node.left)
                    yield from dfs(node.right)
                else:
                    yield from dfs(node.right)
                    yield from dfs(node.left)
                yield '#'
        return all(x==y for x,y in itertools.zip_longest(dfs(root1),dfs(root2)))

TC: O(min(n1,n2)) SC:O(min(H1,H2)) h1,h2 heights of treees.