DFS:101. Symmetric Tree

我另外写了一个参数是两个树结点的函数，判断左边的左子树和右边的右子树，还有左边的右子树和右边的左子树是不是相等的。

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