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problem:
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.
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solution:
//递归解决
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
bool isSymmetric_aux(TreeNode *node1, TreeNode *node2)
{
if(node1== NULL && node2 == NULL)
return true;
if(node1 == NULL || node2 == NULL)
return false;
return node1->val == node2->val && isSymmetric_aux(node1->left, node2->right) && isSymmetric_aux(node1->right, node2->left);
}
bool isSymmetric(TreeNode *root)
{
if(root == NULL)
return true;
return isSymmetric_aux(root->left, root->right);
}