# 101. Symmetric Tree

## Description

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

Example 1:

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


Example 2:

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

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

## Solutions

Solution 1: Recursion

We design a function $dfs(root1, root2)$ to determine whether two binary trees are symmetric. The answer is $dfs(root, root)$.

The logic of the function $dfs(root1, root2)$ is as follows:

• If both $root1$ and $root2$ are null, then the two binary trees are symmetric, return true.
• If only one of $root1$ and $root2$ is null, or if $root1.val \neq root2.val$, then the two binary trees are not symmetric, return false.
• Otherwise, determine whether the left subtree of $root1$ is symmetric to the right subtree of $root2$, and whether the right subtree of $root1$ is symmetric to the left subtree of $root2$. Here we use recursion.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
return dfs(root, root);
}

private boolean dfs(TreeNode root1, TreeNode root2) {
if (root1 == null && root2 == null) {
return true;
}
if (root1 == null || root2 == null || root1.val != root2.val) {
return false;
}
return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
}
}

• /**
* 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 isSymmetric(TreeNode* root) {
function<bool(TreeNode*, TreeNode*)> dfs = [&](TreeNode* root1, TreeNode* root2) -> bool {
if (!root1 && !root2) return true;
if (!root1 || !root2 || root1->val != root2->val) return false;
return dfs(root1->left, root2->right) && dfs(root1->right, root2->left);
};
return dfs(root, root);
}
};

• # 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(root1, root2):
if root1 is None and root2 is None:
return True
if root1 is None or root2 is None or root1.val != root2.val:
return False
return dfs(root1.left, root2.right) and dfs(root1.right, root2.left)

return dfs(root, root)


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func isSymmetric(root *TreeNode) bool {
var dfs func(*TreeNode, *TreeNode) bool
dfs = func(root1, root2 *TreeNode) bool {
if root1 == nil && root2 == nil {
return true
}
if root1 == nil || root2 == nil || root1.Val != root2.Val {
return false
}
return dfs(root1.Left, root2.Right) && dfs(root1.Right, root2.Left)
}
return dfs(root, root)
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

const dfs = (root1: TreeNode | null, root2: TreeNode | null) => {
if (root1 == root2) {
return true;
}
if (root1 == null || root2 == null || root1.val != root2.val) {
return false;
}
return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
};

function isSymmetric(root: TreeNode | null): boolean {
return dfs(root.left, root.right);
}


• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isSymmetric = function (root) {
function dfs(root1, root2) {
if (!root1 && !root2) return true;
if (!root1 || !root2 || root1.val != root2.val) return false;
return dfs(root1.left, root2.right) && dfs(root1.right, root2.left);
}
return dfs(root, root);
};


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root1: &Option<Rc<RefCell<TreeNode>>>, root2: &Option<Rc<RefCell<TreeNode>>>) -> bool {
if root1.is_none() && root2.is_none() {
return true;
}
if root1.is_none() || root2.is_none() {
return false;
}
let node1 = root1.as_ref().unwrap().borrow();
let node2 = root2.as_ref().unwrap().borrow();
node1.val == node2.val &&
Self::dfs(&node1.left, &node2.right) &&
Self::dfs(&node1.right, &node2.left)
}

pub fn is_symmetric(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
let node = root.as_ref().unwrap().borrow();
Self::dfs(&node.left, &node.right)
}
}