算法 101. Symmetric Tree

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) {
        
    }
}

给一个二叉树，检查下是不是自我镜像的（围绕中间对称）。
解：
思路一：递归。递归去判断，当前节点的左节点是否与右节点相同，不同返回 false。
思路二：遍历或者叫迭代。和递归差不多，只不过不是使用递归的方式。
思路三：旋转。这是我一刚开始的想法， 不停递归旋转左右子树，再去判断与原树是否相同，这复杂度还不如直接判断。

以下为代码：
递归方式：

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 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;
}