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