Symmetric Tree

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

两个函数,一个是判断相等,一个是交换左右子树

/**
 * 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 isEqual(TreeNode *root1, TreeNode *root2)
    {
        if(!root1&&!root2)
            return true;
        if(!root1||!root2)
            return false;
        if(!root1->left&&!root1->right&&!root2->left&&!root2->right&&root1->val==root2->val)
            return true;
        if(isEqual(root1->left, root2->left)&&isEqual(root1->right, root2->right)&&root1->val==root2->val)
            return true;
        return false;
    }
    void swapLeftRight(TreeNode *root)
    {
        if(NULL==root||(NULL==root->left&&NULL==root->right))return;
        swapLeftRight(root->left);
        swapLeftRight(root->right);
        TreeNode *temp = root->left;
        root->left = root->right;
        root->right = temp;
    }

    bool isSymmetric(TreeNode *root) {
        if(!root)
            return true;
        swapLeftRight(root->left);
        if(isEqual(root->left,root->right))
            return true;
        return false;
    }
};