【leetcode】Symmetric Tree

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

对称的递归表达式：

testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left);

leftNode->left->val==rightNode->right->val

leftNode->right->val==rightNode->left->val

 1 class Solution {
 2
 3 public:
 4
 5     bool isSymmetric(TreeNode *root) {
 6
 7         if(root==NULL)
 8
 9         {
10
11             return true;
12
13         }
14
15         return testMirror(root->left,root->right);
16
17
18
19     }
20
21
22
23     bool testMirror(TreeNode *leftNode,TreeNode *rightNode)
24
25     {
26
27
28
29         if(leftNode==NULL&&rightNode==NULL)
30
31             return true;
32
33         if(leftNode!=NULL&&rightNode==NULL||leftNode==NULL&&rightNode!=NULL)
34
35             return false;
36
37         if(leftNode->val!=rightNode->val)
38
39             return false;
40
41
42
43         return testMirror(leftNode->left,rightNode->right)&&testMirror(leftNode->right,rightNode->left);
44
45
46
47
48
49     }
50
51 };