Symmetric Tree

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

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

But the following [1,2,2,null,3,null,3] is not:

    1   / \  2   2   \   \   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

• Difficulty: Easy

/** * Given a binary tree, check whether it is a mirror of itself (ie, symmetric * around its center). * @author ning */public class SymmetricTree {public boolean isSymmetric(TreeNode root) {return root == null || isSymmetricHelp(root.left, root.right);}//递归比较左子树的左节点和右子树的右节点以及左子树的右节点和右子树的左节点。private boolean isSymmetricHelp(TreeNode left, TreeNode right) {if (left == null || right == null)return left == right;if (left.val != right.val)return false;return isSymmetricHelp(left.left, right.right) && isSymmetricHelp(left.right, right.left);}}