Prove safety offer58: symmetrical binary tree. Analyzing a binary tree is not symmetric, if a binary image is a binary tree with this same, which is defined as the symmetrical

1 Title Description

　　 Please implement a function, a binary tree is used to determine not symmetrical. Note that if a binary image is a binary tree with this same definition as symmetrical.

2 ideas and methods

　　Define a traversal algorithm, to then traverse right child node traversal left child node; symmetric, preorder: root -> right child node -> the left child node ; and preorder: root - > left child node -> right child node ; by comparison preorder traversal of a binary tree and symmetrical sequence preorder traversal to determine whether the binary tree is the same symmetry. Note meaning: Because the binary elements may all be the same, it will traverse during nullptr also taken into account.

3 C ++ core code

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };
10 */
11 class Solution {
12 public:
13     bool isSymmetrical(TreeNode* pRoot)
14     {
15         return issymmetrical(pRoot,pRoot);
16     }
17     
18     //Comparative binary tree preorder traversal sequence (Root -> Left -> right) and symmetric preorder sequence (Root -> Right -> left), if equal, symmetrical 
. 19      BOOL issymmetrical (the TreeNode * pRoot1, the TreeNode * pRoot2) {
 20 is          
21 is          IF (pRoot1 == NULL && pRoot2 == NULL)
 22 is              return  to true ;         // since there may be a case where all the elements are equal; traversal sequence considered nullptr a
 23 is          
24          // a null pointer, another non-null pointer 
25          IF (pRoot1 == NULL || pRoot2 == NULL)
 26 is              return  to false ;
 27          
28          IF (pRoot1-> Val = pRoot2->! Val)
 29              return  to false ;
 30          
31 is          return issymmetrical(pRoot1->right,pRoot2->left) && issymmetrical(pRoot1->left,pRoot2->right);
32     }
33 };

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Reference material

https://blog.csdn.net/zjwreal/article/details/89282964