Solution
/**
* 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 isSymmetric(TreeNode* root) {
return isSymmetric_r(root->left, root->right);
}
bool isSymmetric_r(TreeNode* l, TreeNode* r) {
if (l == nullptr && r == nullptr) {
return true;
}
if (l == nullptr || r == nullptr) {
return false;
}
bool isL = isSymmetric_r(l->left, r->right);
bool isR = isSymmetric_r(l->right, r->left);
if (l->val == r->val && isL && isR) {
return true;
}
return false;
}
};
Complexity
- Time Complexity: O(N), where N is the number of nodes.
- Space Complexity: O(N), where N is the size of the implicit call stack.
TODO(alex): Do this exercise iteratively.