学步园

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.

递归思路：

对左右子树分别深搜，不过左右子树的优先访问分支是相反的。

递归题解：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool verify(TreeNode* L, TreeNode* R)
    {
        if (L == nullptr && R == nullptr)
            return true;
        else if (bool(L) ^ bool(R))
            return false;
        else if (L->val != R->val)
            return false;
        else 
            return verify(L->right, R->left) && verify(L->left, R->right);
    }

    bool isSymmetric(TreeNode *root) {
        if (root == nullptr)
            return true;
        
        return verify(root->left, root->right);
    }
};

迭代思路：

按照行遍历，每一行都要考虑到有nullptr的可能，直到这一行的值都是nullptr为止。