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:



But the following is not:



Note:

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.

分析

判断一棵二叉树是否为对称树;

仍然采用递归的思想,判断该树的左右子树是否对称;

若二叉树p与二叉树q对称,也就是说其根节点相同,p左子树应与q右子树对称,同理,p右子树应与q左子树对称;

AC代码

/**
* 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 isSymmetric(TreeNode* root) {
if (!root)
return true;
else
//判断左右子树是否对称
return isSymmetricTree(root->left, root->right);
} bool isSymmetricTree(TreeNode* p, TreeNode* q) {
//如果两个二叉树均为空,则返回true
if (!p && !q)
{
return true;
}
//如果两者其一为空树,则返回false
else if (!p || !q)
{
return false;
}
else{
if (p->val != q->val)
return false;
else
//p左子树应与q右子树对称,同理,p右子树应与q左子树对称
return isSymmetricTree(p->left, q->right) && isSymmetricTree(p->right, q->left);
}
}
};

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