[leetcode] 10. Symmetric Tree
这次我觉得我的智商太低,想了很久才写出来。题目是让求镜像二叉树判断,题目如下:
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 3Note:
Bonus points if you could solve it both recursively and iteratively.confused what
"{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1
/ \
2 3
/
4
\
5The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}"
.
我之前的思路有问题,我是想先做一个函数然后记录到某个子叶的路径,这个路径拿左右来表示,0表示左,1表示右。结果这个函数的递归的版本我没写出来,因为结束弹出方法我实在没找到,然后迭代我觉得实在太烦了。然后就想着换另一种方法:把整个函数的出口另外做一个简单的递归,然后在主函数里面调用一次就行。
题解如下:
/**
* 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 Symmetric(TreeNode *left, TreeNode *right)
{
if (left == NULL && right == NULL)
{
return true;
}
if (left == NULL || right == NULL)
{
return false;
}
return left->val == right->val && Symmetric(left->left, right->right) && Symmetric(left->right, right->left);
} bool isSymmetric(TreeNode *root)
{
return root != NULL ? Symmetric(root->left, root->right) : true;
}
};
即如上。之后leetcode的题目好像要买它的书之后才能再刷了,不过那是对应的是新出的题目,老的题目还是可以的,先把之前的题目刷完再说吧。
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