[LeetCode] 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.

Note:子树不需要对称,只需要根书的左右是镜像即可。

1 递归,判断leftNode->left和rightNode->right,leftNode->right和rightNode->left,left->val == right->val

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11     public:
12         bool isSymmetric(TreeNode *root) {
13             if(root == NULL) return true;
14             return isSymmetricTree(root->left, root->right);
15         }
16 
17         bool isSymmetricTree(TreeNode *left, TreeNode *right)
18         {
19             if(left == NULL && right == NULL) return true; //终止条件
20             if(left == NULL || right == NULL) return false;
21             if(left->val == right->val &&          //合并
22                     isSymmetricTree(left->left, right->right) &&
23                     isSymmetricTree(right->left, left->right) 
24               )   
25                 return true;
26 
27             return false;
28         }
29 
30 };
原文地址:https://www.cnblogs.com/diegodu/p/3780446.html