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.

tag: 经典binary tree

recursion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        return isMirror(root, root);
    }

    private boolean isMirror(TreeNode root1, TreeNode root2){
        if (root1 == null && root2 == null) return true;
        if (root1 == null || root2 == null) return false;

        return root1.val == root2.val && isMirror(root1.left, root2.right) && isMirror(root1.right, root2.left);
    }
}

Iterative

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<>();

        q.offer(root);
        q.offer(root);

        while (!q.isEmpty()){
            TreeNode n1 = q.poll();
            TreeNode n2 = q.poll();

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

        return true;
    }
}