AcWing 39. 对称的二叉树
原题链接
简单
作者:
回归线
,
2021-01-31 02:28:16
,
所有人可见
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阅读 347
/**
* 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) {
return true;
}
return isEqual(root->left, root->right);
}
bool isEqual(TreeNode* a, TreeNode* b) {
if (!a || !b) {
return !a && !b;
}
return a->val == b->val && isEqual(a->left, b->right) && isEqual(a->right, b->left);
}
};