标签：right TreeNode val nullptr 二叉树 return 101 LeetCode left
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool dfs(TreeNode* left, TreeNode* right) {
if (left == nullptr && right != nullptr) return false;
else if (left != nullptr && right == nullptr) return false;
else if (left == nullptr && right == nullptr) return true;
else if (left->val != right->val) return false;
//当外侧或内侧值相等时才进行下一层比较
bool outside = dfs(left->left, right->right);
bool inside = dfs(left->right, right->left);
return outside && inside;
}
bool isSymmetric(TreeNode* root) {
if (root == nullptr) return true;
return dfs(root->left, root->right);
}
};
标签：right,
TreeNode,
val,
nullptr,
二叉树,
return,
101,
LeetCode,
left
From： https://www.cnblogs.com/hjy94wo/p/16664052.html