Description
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?
Code
recursive
/**
* 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 !root || helper(root->left, root->right);
}
bool helper(TreeNode* left, TreeNode* right) {
if(!left || !right) return left == right;
if(left->val != right->val) return false;
return helper(left->right, right->left) && helper(left->left, right->right);
}
};Iterative
using queue.
/**
* 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;
queue<TreeNode*> check;
check.push(root->left);
check.push(root->right);
while (!check.empty()) {
TreeNode* node1 = check.front();
check.pop();
TreeNode* node2 = check.front();
check.pop();
if (!node1 && node2) return false;
if (!node2 && node1) return false;
if (node1 && node2) {
if (node1->val != node2->val) return false;
check.push(node1->left);
check.push(node2->right);
check.push(node1->right);
check.push(node2->left);
}
}
return true;
}
};