Symmetric tree <leetcode>

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

For example, this binary tree is unsupported Ric:

 

    1   /   2   2 / \ / 3  4 4  3

 

But the following is not:

1/2 2 \ 3 3


Idea: first, we will use recursion to traverse the Left and Right sub-trees. When different sub-trees are traversed, We will traverse the root node first, then the left child, and finally the right child; when traversing the right subtree, the county traverses the root node, then the right child, and finally the left child, and compares each traversal to a node. The Code is as follows: // No optimization

 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         TreeNode *rootl;15         TreeNode *rootr;16         rootl=root->left;17         rootr=root->right;18         return isSame(rootl,rootr);19     }20     21     bool isSame(TreeNode *rootl,TreeNode *rootr)22     {23         if(rootl==NULL||rootr==NULL)24         {25             if(rootl!=NULL||rootr!=NULL)26             {27                 return false;28             }29             else return true;30         }31         32         if(rootl->val!=rootr->val)  return false;33         else34         {35             if(isSame(rootl->left,rootr->right))36             {37                 if(isSame(rootl->right,rootr->left))38                 {39                     return true;40                 }41                 else return false;42             }43             else return false;44         }45         46     }47 };

 

Symmetric tree <leetcode>