Symmetric Binary Tree

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

Example

    1
   / \
  2   2
 / \ / \
3  4 4  3

is a symmetric binary tree.

    1
   / \
  2   2
   \   \
   3    3

is not a symmetric binary tree.

Solution: DFS

public boolean isSymmetric(TreeNode root) {
    if(root == null) {
        return true;
    }
    return isSymmetric(root.left, root.right);
}

public boolean isSymmetric(TreeNode left, TreeNode right) {
    if(left == null && right == null) {
        return true;
    }
    if(left == null || right == null) {
        return false;
    }
    return left.val == right.val && isSymmetric(left.left, right.right) &&
isSymmetric(left.right, right.left);
}