LeetCode 솔루션 분류
[Easy - wk4 - Q2] 101. Symmetric Tree
본문
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:
Input: root = [1,2,2,3,4,4,3] Output: true
Example 2:
Input: root = [1,2,2,null,3,null,3] Output: false
Constraints:
- The number of nodes in the tree is in the range
[1, 1000]. -100 <= Node.val <= 100
Follow up: Could you solve it both recursively and iteratively?
관련자료
댓글 2
mingki님의 댓글
- 익명
- 작성일
C++
Runtime: 14 ms, faster than 13.66% of C++ online submissions for Symmetric Tree.
Memory Usage: 16.3 MB, less than 93.41% of C++ online submissions for Symmetric Tree.
Runtime: 14 ms, faster than 13.66% of C++ online submissions for Symmetric Tree.
Memory Usage: 16.3 MB, less than 93.41% of C++ online submissions for Symmetric Tree.
class Solution {
bool helper(TreeNode *left, TreeNode *right) {
if (!left && !right) return true;
if (!left || !right) return false;
return left->val == right->val &&
helper(left->left, right->right) &&
helper(left->right, right->left);
}
public:
bool isSymmetric(TreeNode *root) {
return helper(root, root);
}
};dawn27님의 댓글
- 익명
- 작성일
JS
Runtime: 77 ms, faster than 79.08% of JavaScript online submissions for Symmetric Tree.
Memory Usage: 44.3 MB, less than 85.73% of JavaScript online submissions for Symmetric Tree.
Runtime: 77 ms, faster than 79.08% of JavaScript online submissions for Symmetric Tree.
Memory Usage: 44.3 MB, less than 85.73% of JavaScript online submissions for Symmetric Tree.
var isSymmetric = function(root) {
if (root === null) {
return true;
}
return isMirror(root.left, root.right);
}
var isMirror = function(tree1, tree2) {
if (tree1 === null || tree2 === null) {
return tree1 === tree2;
}
if (tree1.val !== tree2.val) {
return false;
}
return isMirror(tree1.left, tree2.right) && isMirror(tree1.right, tree2.left);
};
