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.
题目描述:
对称树的判断。
代码实现:
/**
* 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 root ? isSymmetric(root->left,root->right) : true;
}
bool isSymmetric(TreeNode *left,TreeNode *right)
{
if(!left && !right) return true;//终止条件
if(!left || !right) return false;//终止条件
return left->val== right->val && isSymmetric(left->left,right->right) && isSymmetric(left->right,right->left);//三方合并
}
};