LeetCode 101. Symmetric Tree

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

/**
 * 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 *lhs, TreeNode *rhs)
    {
        if (NULL == lhs&&NULL == rhs)return true;
        if (NULL!=lhs&&NULL!=rhs&&lhs->val == rhs->val)
        {
            return isSymmetric(lhs->right, rhs->left) && isSymmetric(lhs->left, rhs->right);
        }
        return false;
    }
    bool isSymmetric(TreeNode* root) {
        if (!root)return true;
        return isSymmetric(root->left, root->right);
    }
};
原文地址:https://www.cnblogs.com/csudanli/p/5877084.html