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

/**
 * 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) {
        
    }
}

給一個二叉樹，檢查下是不是自我鏡像的（圍繞中間對稱）。
解：
思路一：遞歸。遞歸去判斷，當前節(jié)點的左節(jié)點是否與右節(jié)點相同，不同返回 false。
思路二：遍歷或者叫迭代。和遞歸差不多，只不過不是使用遞歸的方式。
思路三：旋轉。這是我一剛開始的想法， 不停遞歸旋轉左右子樹，再去判斷與原樹是否相同，這復雜度還不如直接判斷。

以下為代碼：
遞歸方式：

public boolean isSymmetric(TreeNode root) {
    return isMirror(root, root);
}

public boolean isMirror(TreeNode t1, TreeNode t2) {
    if (t1 == null && t2 == null) {
        return true;
    }
    if (t1 == null || t2 == null) {
        return false;
    }
    return (t1.val == t2.val)
        && isMirror(t1.right, t2.left)
            && isMirror(t1.left, t2.right);
}

迭代方式：

public boolean isSymmetric(TreeNode root) {
    Queue<TreeNode> q = new LinkedList<>();
    q.add(root);
    q.add(root);
    while (!q.isEmpty()) {
        TreeNode t1 = q.poll();
        TreeNode t2 = q.poll();
        if (t1 == null && t2 == null) continue;
        if (t1 == null || t2 == null) return false;
        if (t1.val != t2.val) return false;
        q.add(t1.left);
        q.add(t2.right);
        q.add(t1.right);
        q.add(t2.left);
    }
    return true;
}