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);

        }

    }

};