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:
[1, 1000]
.-100 <= Node.val <= 100
Follow up: Could you solve it both recursively and iteratively?
using LeetCodeNet.Com_github_leetcode;
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public bool IsSymmetric(TreeNode root) {
if (root == null) {
return true;
}
return Helper(root.left, root.right);
}
private 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);
}
}