101. 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.

public class Solution {
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);
}
}

public boolean isSymmetric(TreeNode root) {
if(root == null)
return true;
if(root.left == null && root.right == null)
return true;
if(root.left == null || root.right == null)
return false;
LinkedList<TreeNode> q1 = new LinkedList<TreeNode>();
LinkedList<TreeNode> q2 = new LinkedList<TreeNode>();
q1.add(root.left);
q2.add(root.right);
while(!q1.isEmpty() && !q2.isEmpty()){
TreeNode n1 = q1.poll();
TreeNode n2 = q2.poll();

if(n1.val != n2.val)
return false;
if((n1.left == null && n2.right != null) || (n1.left != null && n2.right == null))
return false;
if((n1.right == null && n2.left != null) || (n1.right != null && n2.left == null))
return false;

if(n1.left != null && n2.right != null){
q1.add(n1.left);
q2.add(n2.right);
}

if(n1.right != null && n2.left != null){
q1.add(n1.right);
q2.add(n2.left);
}
}
return true;
}