[leetcode-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.

思路：

递归法：当要比较的两个结点有一个为空时返回false，当要比较的两个结点值不相等时返回false。 当两结点为空时即为true。

非递归：层次遍历，需要注意的是左结点按照左右的顺序进入队列，而右边结点要按照先右后左顺序进入队列，这样比较的时候才满足是否对称要求。

bool isSymmetric2(TreeNode *root)
    {//非递归 层次遍历二叉树
        TreeNode *left, *right;
        if (!root)
            return true;
        queue<TreeNode*> q1, q2;
        q1.push(root->left);
        q2.push(root->right);
        while (!q1.empty() && !q2.empty()){
            left = q1.front();
            q1.pop();
            right = q2.front();
            q2.pop();
            if (NULL == left && NULL == right)
                continue;
            if (NULL == left || NULL == right)
                return false;
            if (left->val != right->val)
                return false;
            q1.push(left->left);
            q1.push(left->right);
            q2.push(right->right);
            q2.push(right->left);
        }
        return true;
    }
    bool isSymmetricCore(TreeNode* left, TreeNode* right)
    {
        if (left == NULL && right == NULL) return true;
        if (left == NULL && right != NULL || left != NULL && right == NULL || left->val != right->val)return false;
        return isSymmetricCore(left->left, right->right) && isSymmetricCore(left->right, right->left);//递归判断
    }
    bool isSymmetric(TreeNode* root)
    {
        if (root == NULL) return true;
        return isSymmetricCore(root->left, root->right);
    }

参考：

https://discuss.leetcode.com/topic/4332/my-c-accepted-code-in-16ms-with-iteration-solution