101. Symmetric Tree

# Easy

Be careful with edge case in compare() function.

1. root1 = None & roo2 = None

2. root1 = None & root2 != None

3. root1 != None & roo2 = None

Python

# 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 isSymmetric(self, root: TreeNode) -> bool:
        # edge case
        if root == None:
            return True
        
        # regular case
        root1 = root.left
        root2 = root.right
        return self.compare(root1, root2)
            
    def compare(self, root1, root2):
        # stop condition
        if root1 == None and root2 == None:
            return True
        if (root1 != None and root2 == None) or (root1 == None and root2 != None):
            return False
        
        # regular case
        if root1.val == root2.val:
            a1 = self.compare(root1.left, root2.right)
            a2 = self.compare(root1.right, root2.left)
            if a1 == True and a2 == True:
                return True
        else:
            return False

Second Tab