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.

难度：easy

题目：给定二叉树检查它是否为镜像。

思路：同same Tree

Runtime: 7 ms, faster than 46.76% of Java online submissions for Symmetric Tree.
Memory Usage: 30.1 MB, less than 0.94% of Java online submissions for Symmetric Tree.

/**
 * 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 (null == root) {
            return true;
        }
        return isSymmetric(root.left, root.right);
    }
    
    public boolean isSymmetric(TreeNode p, TreeNode q) {
        if (p == null && q != null || p != null && q == null) {
            return false;
        }
        if (p == null && q == null) {
            return true;
        }
        if (p.val != q.val) {
            return false;
        }
        
        return isSymmetric(p.left, q.right) && isSymmetric(p.right, q.left);
    }
}