Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1   /   2   2 / \ / 3  4 4  3

But the following is not:

    1   /   2   2   \      3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

这题就是深搜，但是要同时搜两颗树，左子树和右子树。然后因为要是对称的，所以左子树要先搜左边，右子树要先搜右边，然后再反过来

这题说用迭代写出来可以bonus，暂时没写出来，不过也不会太难，这题迭代的深搜应该用两个栈吧，一个左栈一个右栈...(猜的

bool symmetricIter(TreeNode *left, TreeNode *right) {    if (!left && !right) return true;    if (!(left && right)) return false;    if (left->val == right->val) {        return symmetricIter(left->left, right->right) && symmetricIter(left->right, right->left);    }    return false;}bool isSymmetric(TreeNode *root) {    return symmetricIter(root->left, root->right);}

[LeetCode] Symmetric Tree