101. Symmetric Tree
一、问题描述
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
二、输入输出
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/
\
2 2
/
\ /
\
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/
\
2 2
\ \
3 3
Note: Bonus points if you could solve it both recursively and iteratively.
三、解题思路
递归遍历
这道题让你判断一颗树是不是对称的。递归方法会加分。其实递归也不难，但是需要你仔细观察。从根节点开始，递归遍历，看左子树和右子树是否对称。然后继续遍历左字树，出问题了，这时候需要右子树的信息来判断是否是对称的。所以函数的接口一定是两个TreeNode的。我们需要判断leftTree->left isSymmetric rightTree->right && leftTree->right isSysmmetirc rightTree->left 必须判断这两个才行。非常的新鲜写法。另外，单独处理两个root是否为空或者其中一个为空的写法的时候，下面这种写法值得借鉴：
if(left ==
nullptr || right ==
nullptr){
return (left == right);
}
||表示其中一个为空或者两个都为空
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool
isSymmetricRecusive(
const TreeNode* left,
const TreeNode* right)
{
if(left == nullptr || right == nullptr){
return (left == right);
}
if(left->val != right->val)
return false;
return isSymmetricRecusive(left->left, right->right) && isSymmetricRecusive(left->right, right->left);
}
bool isSymmetric(TreeNode* root) {
return (root == nullptr || isSymmetricRecusive(root->left, root->right));
}
};
