---
tags: data_structure_python
---

# Symmetric Tree <img src="https://img.shields.io/badge/-easy-brightgreen">

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

## Solution

### Solution 1: Depth First Search

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

class Solution:
    def __isSymmetric(self, root1, root2):
        if root1 is None and root2 is None:
            return True
        elif root1 is None or root2 is None:
            return False
        else:
            return ((root1.val == root2.val) 
                and self.__isSymmetric(root1.left, root2.right) 
                and self.__isSymmetric(root1.right, root2.left))
    
    def isSymmetric(self, root: TreeNode) -> bool:
        return self.__isSymmetric(root, root)

### Solution 2: Breadth First Search

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

class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
        stack = []
        stack.append(root)
        stack.append(root)

        while len(stack) > 0:
            isNotSkip = True
        
            node1 = stack.pop(0)
            node2 = stack.pop(0)
            
            if node1 is None and node2 is None:
                isNotSkip = False
            elif node1 is None or node2 is None:
                return False
            elif node1.val != node2.val:
                return False
            
            if isNotSkip:
                stack.append(node1.left)
                stack.append(node2.right)

                stack.append(node1.right)
                stack.append(node2.left)

        return True