LeetCode-in-All
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:
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?
Solution
/**
* Definition for a binary tree node.
* class TreeNode {
* int val;
* TreeNode? left;
* TreeNode? right;
* TreeNode([this.val = 0, this.left, this.right]);
* }
*/
class Solution {
bool isSymmetric(TreeNode? root) {
if (root == null) {
return true;
}
return _helper(root.left, root.right);
}
bool _helper(TreeNode? leftNode, TreeNode? rightNode) {
if (leftNode == null || rightNode == null) {
return leftNode == null && rightNode == null;
}
if (leftNode.val != rightNode.val) {
return false;
}
return _helper(leftNode.left, rightNode.right) && _helper(leftNode.right, rightNode.left);
}
}