101. Symmetric Tree

目录

• 题目描述
• 代码实现

题目描述

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.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

代码实现

class Solution {
public:
    bool isSymmetric(TreeNode *root) {
      return isSymmetricCore(root,root);
        
    }
    bool isSymmetricCore(TreeNode* root1,TreeNode* root2){
      if(root1 == nullptr && root2 == nullptr)
        return true;
      if(root1 == nullptr || root2 == nullptr)
        return false;
      if(root1->val != root2->val)
        return false;
      return isSymmetricCore(root1->left,root2->right) && isSymmetricCore(root1->right,root2->left);

    }
};