- Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
/**
* 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) {
queue<TreeNode*> q;
q.push(root);
q.push(root);
while(!q.empty()){
TreeNode* t1 = q.front();
q.pop();
TreeNode* t2 = q.front();
q.pop();
if(t1 == nullptr && t2 == nullptr) continue;
if(t1 == nullptr || t2 == nullptr) return false;
if(t1->val != t2->val) return false;
q.push(t1->left);
q.push(t2->right);
q.push(t1->right);
q.push(t2->left);
}
return true;
}
};
/**
* 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) {
return is_mirrow(root, root);
}
bool is_mirrow(TreeNode* t1, TreeNode* t2){
if(t1 == nullptr && t2 == nullptr) return true;
if(t1 == nullptr || t2 == nullptr) return false;
return (t1->val == t2->val) && is_mirrow(t1->left, t2->right) && is_mirrow(t1->right, t2->left);
}
};
