101. Symmetric Tree 对称树,递归算法

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

Note:
Bonus points if you could solve it both recursively and iteratively.

基本思想：对称树左右子树相互对称。如果互换两个子树的右分支，两个子树还是对称树。

我的答案：（C++）

/**
 * 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==NULL)return true;
        else if(root->left==NULL&&root->right==NULL)
            return true;
        else if(root->left!=NULL&&root->right!=NULL)
        {
            if(root->left->val==root->right->val){
                TreeNode *temp=root->left->right;
                root->left->right=root->right->right;
                root->right->right=temp;
                return isSymmetric(root->left)&&isSymmetric(root->right);
            }
            return false;
        }
        return false;
    }
};