# 101. Symmetric Tree

#### Easy

***

Given the `root` of a binary tree, *check whether it is a mirror of itself* (i.e., symmetric around its center).

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/02/19/symtree1.jpg)

```
Input: root = [1,2,2,3,4,4,3]
Output: true
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/02/19/symtree2.jpg)

```
Input: root = [1,2,2,null,3,null,3]
Output: false
```

 

**Constraints:**

* The number of nodes in the tree is in the range `[1, 1000]`.
* `-100 <= Node.val <= 100`

&#x20;

**Follow up:** Could you solve it both recursively and iteratively?

#### Recursion

```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: Optional[TreeNode]) -> bool:
        if root is None:
            return True
        def recursion(left, right):
            if left is None and right is None:
                return True
            elif left is None or right is None:
                return False
            elif left.val != right.val:
                return False
            return recursion(left.right, right.left) and recursion(left.left, right.right)
        return recursion(root.left, root.right)
            
```

#### Iterative

```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: Optional[TreeNode]) -> bool:
        # .
        if root is None:
            return True
        q = [[root.left, root.right]]
        while q:
            left, right = q.pop(0)
            if left is None and right is None:
                continue
            elif left is None or right is None:
                return False
            elif left.val != right.val:
                return False
            else:
                q.append([left.right, right.left])
                q.append([left.left, right.right])
        return True
            
```