从简单的道题目開始刷题目:

Symmetric Tree

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 3

But 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);
}
};

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