递归判断树是否是对称的，不难，但是要注意root为NULL和某一个结点为NULL的情况
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode* node1, TreeNode* node2) {
if(node1 == NULL && node2 == NULL)
return true;
if(node1 == NULL || node2 == NULL)
return false;
if(node1->val == node2->val &&
isSymmetric(node1->left, node2->right) &&
isSymmetric(node1->right, node2->left)) {
return true;
}
return false;
}
bool isSymmetric(TreeNode* root) {
if(root == NULL)
return true;
return isSymmetric(root->left, root->right);
}
};
