# 101 Symmetric Tree – Easy

### Problem:

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.

### Thoughts:

This is very similar to the Same Tree problem. Very straight forward in recursion way.

### Solutions:

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
// return recursion(root.left, root.right);
return iterative(root);
}
private boolean recursion(TreeNode left, TreeNode right) {
if (left == null && right == null) {
return true;
}
if (left != null && right != null && left.val == right.val) {
return recursion(left.left, right.right) && recursion(left.right, right.left);
}
return false;
}
private boolean iterative(TreeNode root) {
Stack<TreeNode> left = new Stack<TreeNode>();
Stack<TreeNode> right = new Stack<TreeNode>();
left.push(root.left);
right.push(root.right);
while (left.size() > 0) {
TreeNode l = left.pop();
TreeNode r = right.pop();
if (l == null && r == null) {
continue;
}
if (l != null && r != null && l.val == r.val) {
left.push(l.left);
right.push(r.right);
left.push(l.right);
right.push(r.left);
continue;
}
return false;
}
return true;
}
}
``````