Leetcode 101. Symmetric Tree

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

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

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.

解法：判断一棵树是不是对称的，就判断对应的两个结点的val是不是一样，首先是根的左右两个结点，之后是左字数的右子数和右字数的左字数，。。。。以此类推