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

Example 1:

                 1

           2 2

    3 4 4 3

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

Output: true

Example 2:

                 1

           2 2

    null 3 null 3

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

/**

 * 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) {

        List<Integer> serialized = new ArrayList<Integer>();

        InorderTraversal(root, serialized);

        return IsPalindrome(serialized);

    }

    public boolean isPalindrome(List<Integer> serialized) {

        int i = 0;

        int j = serialized.count() - 1;

        while (i < j) {

            if (serialized.getAt(i) != serialized.getAt(j)) {

                return false;

            }

            i++;

            j--;

        }

        return true;

    }

    public void InorderTraversal(TreeNode root, List<Imteger> serialized) {

        if (root == null) {

            serialized.Add(Integer.MinValue);

            return;

        }

        InOrderTraversal(root.left);

        serialized.Add(root.val);

        InOrderTraversal(root.right);

    }

}

#blogpost: https://1drv.ms/w/c/d609fb70e39b65c8/EcVxcTYDkE9Hro_eEthnuH8Bx-PIXIxbdNq2kXKuyr8TdA?e=a1ofbg