问题:判断二叉树是否为镜像二叉树 分析:递归判断,根节点单独判断,然后递归左结点和右结点,之后每次一起递归左结点的左结点和右结点的右结点比较,左结点的右结点和右结点的左结点比较
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool dfs(TreeNode *left,TreeNode *right)
{
if(!left && !right) return true;//需要先判断这个,不然val那个可能会出现RE
if((left && !right) || (!left && right) || (left->val!=right->val)) return false;
return dfs(left->left,right->right) && dfs(left->right,right->left);
}
bool isSymmetric(TreeNode *root) {
if(root==NULL || (!root->left && !root->right)) return true;
return dfs(root->left,root->right);
}
};