CODE 32: Symmetric Tree

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

For example, this binary tree is symmetric:

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

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as  "{1,2,3,#,#,4,#,#,5}".

	public boolean isSymmetric(TreeNode root) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if (null == root) {
			return true;
		}
		return dfs(root.left, root.right);
	}

	boolean dfs(TreeNode left, TreeNode right) {
		if ((null != left && null == right) || (null == left)
				&& (null != right)) {
			return false;
		} else if (left == null && right == null) {
			return true;
		} else if (left.val != right.val) {
			return false;
		}
		boolean leftResult = dfs(left.left, right.right);
		boolean rightResult = dfs(left.right, right.left);

		return (leftResult && rightResult) ? true : false;
	}