951 Flip Equivalent Binary Trees

Link: https://leetcode.com/problems/flip-equivalent-binary-trees/ 

Thought

If 2 nodes are equivalent, either their left and right side match, or they are flipped match. So using a recursion(why recursion? It is a Tree!), and try to match each node pairs from root1 and root2.

For each recursion level, I will first compare if current node pair have the same value. (I can return False immediately if not match). If it matches, there are 3 cases:

1. They are identical -> recursion with (root1.left, root2.left,) and  (root1.right, root2.right) should return True
2. They are flipped -> recursion with (root1.right, root2.left,) and  (root1.left, root2.right) should return True
   
3. They are different -> Otherwise

Code

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def flipEquiv(self, root1, root2):
        """
        :type root1: TreeNode
        :type root2: TreeNode
        :rtype: bool
        """
        def compare(node1,node2):
            if node1 == None and node2 == None:
                return True
            if node1 == None or node2 == None:
                return False
            if node1.val != node2.val:
                return False
            res = compare(node1.left, node2.left) and compare(node1.right, node2.right)
            res = res or compare(node1.left, node2.right) and compare(node1.right, node2.left)
            return res
        return compare(root1, root2)

Runtime

Time: O(n)

Space: O(1)