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 3 4 5 6 |
1 / \ 2 2 / \ / \ 3 4 4 3 |

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

1 2 3 4 5 6 |
1 / \ 2 2 \ \ 3 3 |

Note:

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

Analysis

This problem can be easily solved using a recursive algorithm. We defined a method with the following signature:

equal(left, right)

then we recursively call equal(left.left, right.right) and equal(left.right, right.left).

The method returns false as long as only one of the two nodes is null,

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