---
# System prepended metadata

title: LC 101. Symmetric Tree
tags: [easy, Binary Tree, leedcode, python]

---

# LC 101. Symmetric Tree

### [Problem link](https://leetcode.com/problems/symmetric-tree/)

###### tags: `leedcode` `python` `easy` `Binary Tree`

Given the <code>root</code> of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

**Example 1:** 
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/symtree1.jpg" style="width: 354px; height: 291px;" />
```
Input: root = [1,2,2,3,4,4,3]
Output: true
```

**Example 2:** 
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/symtree2.jpg" style="width: 308px; height: 258px;" />
```
Input: root = [1,2,2,null,3,null,3]
Output: false
```

 **Constraints:** 

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


 **Follow up:**  Could you solve it both recursively and iteratively?
## Solution 1 - DFS (recursively)
#### Python
```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:
        def dfs(left, right):
            if left and right and left.val == right.val:
                return dfs(left.left, right.right) and dfs(left.right, right.left)

            return left == right

        return dfs(root.left, root.right)
```
#### C++
```cpp=
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool dfs(TreeNode *left, TreeNode *right) {
        if (left == nullptr || right == nullptr) {
            return left == right;
        }
        return left->val == right->val && dfs(left->left, right->right) && dfs(left->right, right->left);
    }

    bool isSymmetric(TreeNode* root) {
        return dfs(root->left, root->right);
    }
};
```

## Solution 2 - DFS (iteratively)
#### Python
```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:
        q = deque([root.left, root.right])

        while q:
            left = q.pop()
            right = q.pop()

            if not left and not right:
                continue

            if not left or not right or left.val != right.val:
                return False

            q.append(left.left)
            q.append(right.right)
            q.append(left.right)
            q.append(right.left)

        return True
```
#### C++
```cpp=
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        queue<TreeNode *> q;

        q.push(root->left);
        q.push(root->right);
        while (!q.empty()) {
            TreeNode *left = q.front();
            q.pop();
            TreeNode *right = q.front();
            q.pop();

            if (left == nullptr && right == nullptr) {
                continue;
            }
            if (left == nullptr || right == nullptr || (left->val != right->val)) {
                return false;
            }
            q.push(left->left);
            q.push(right->right);
            q.push(left->right);
            q.push(right->left);
        }
        return true;
    }
};
```

>### Complexity
>|             | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1  | O(n)            | O(n)             |
>| Solution 2  | O(n)            | O(n)             |


## Note
[類似LC 100. Same Tree](https://hackmd.io/@Alone0506/BkJywty3h)
