Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / 2 2 / / 3 4 4 3
But the following is not:
1 / 2 2 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
bool isSameTree(TreeNode *p, TreeNode *q) { if(p == NULL || q == NULL) return p == q; else return (p->val == q->val) && isSameTree(p->left, q->right) && isSameTree(p->right,q->left); } bool isSymmetric(TreeNode *root) { if(root == NULL) return true; if(root->left == NULL || root->right == NULL) return root->left == root->right;return isSameTree(root->left,root->right); }