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);        }    }};