//101
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.
class Solution {
public:
bool isSymmetric(TreeNode* root) {
if(root==NULL){
return true;
}
return isMirrow(root->left,root->right);
}
private:
//aux functions
bool isMirrow(TreeNode* p,TreeNode*q){
if(p==NULL && q==NULL){
return true;
}
if(p==NULL || q==NULL){
return false;
}
if(p->val==q->val){
bool left=isMirrow(p->left,q->right);
bool right=isMirrow(p->right,q->left);
return left && right;
} else{
return false;
}
}
};
