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

思路：設兩個遊標p,q同時遞歸，遞歸的方向相反。即當p往左走時，q往右走。p往右走時，q往左走。
如果這棵樹是鏡像的，那麽最後 p 和 q一定是同時到達null位置，並且在這個過程中，p,q節點的
值是相同的。

/**
 * 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) {
        if (root==null)
            return true;
        return
 
 judge(root.left,root.right);
    }

    private  boolean judge(TreeNode p,TreeNode q){
        if (p==null&&q==null)               //同時到達null位置，說明樹是鏡像的
            return true;
        if (p==null&&q!=null)
            return false;
        if (p!=null&&q==null)
            return
 
 false;
        if (p.val!=q.val)
            return false;
        return judge(p.left,q.right)&&judge(p.right,q.left);
    }
}

[Leetcode]101. Symmetric Tree