标签:mil size art eth pad otto sel bsp ott
从简单的道题目開始刷题目:
题目: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 3But the following is not:
1 / 2 2 \ 3 3
题目分析:
第一道题目简单的题目,主要利用递归方法,保存左右两个结点。对于LeftNode结点和RightNode结点,推断LeftNode的左结点和RightNode的右结点和LeftNode的右结点和RightNode的左结点是否相等就可以,仅仅要有不相等就能够结束。跳出递归。
代码:
/**
* 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 check(TreeNode *leftNode,TreeNode *rightNode)
{
if(leftNode == NULL && rightNode == NULL)
return true;
if(leftNode == NULL ||rightNode ==NULL)
return false;
if(leftNode ->val != rightNode->val)
return false;
return check(leftNode->left,rightNode->right) && check(leftNode->right,rightNode->left);
}
bool isSymmetric(TreeNode *root) {
if(root == NULL)
return true;
return check(root->left,root->right);
}
};标签:mil size art eth pad otto sel bsp ott
原文地址:http://www.cnblogs.com/lytwajue/p/6953280.html