[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.

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

 

Hide Tags

  Tree Depth-first Search

 

---

　　这个嘛，用两个队列，一个维护树从左到右，一个维护树的从右到左。这就好了。写了两个，一个识queue 中是已经已经符合的，弹出后判断其子节点是否符合，这样queue 中不维护NULL。

 1 class Solution {

 2 public:

 3     bool isSymmetric(TreeNode *root) {

 4         if(root==NULL||(root->left==NULL&&root->right==NULL))  return true;

 5         if(root->left==NULL||root->right==NULL||root->left->val!=root->right->val)   return false;

 6         queue<TreeNode*> lft;

 7         lft.push(root->left);

 8         queue<TreeNode*> rgt;

 9         rgt.push(root->right);

10         while(( !lft.empty() )||( !rgt.empty() )){

11             int nlft=lft.size(),nrgt=rgt.size();

12             if(nlft!=nrgt)  return false;

13             for(int i=0;i<nlft;i++){

14                 TreeNode* curlft=lft.front(),*currgt=rgt.front();

15                 if( (curlft->left==NULL&&currgt->right!=NULL) ||

16                     (curlft->right==NULL&&currgt->left!=NULL) ) return false;

17                 if(curlft->left!=NULL&&

18                    currgt->right!=NULL&&

19                    curlft->left->val!=currgt->right->val)   return false;

20                 if(curlft->right!=NULL&&

21                    currgt->left!=NULL&&

22                    curlft->right->val!=currgt->left->val)   return false;

23                 if(curlft->left!=NULL)  lft.push(curlft->left);

24                 if(curlft->right!=NULL) lft.push(curlft->right);

25                 if(currgt->right!=NULL) rgt.push(currgt->right);

26                 if(currgt->left!=NULL)  rgt.push(currgt->left);

27                 lft.pop();

28                 rgt.pop();

29             }

30         }

31         return true;

32     }

33 };

View Code

 

　　一个是queue 中维护未判断的，包括NULL，这样写逻辑比较简单。

 1 #include <iostream>

 2 #include <queue>

 3 using namespace std;

 4 /**

 5  * Definition for binary tree

 6  */

 7 struct TreeNode {

 8     int val;

 9     TreeNode *left;

10     TreeNode *right;

11     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

12 };

13 /**

14 class Solution {

15 public:

16     bool isSymmetric(TreeNode *root) {

17         if(root==NULL||(root->left==NULL&&root->right==NULL))  return true;

18         if(root->left==NULL||root->right==NULL||root->left->val!=root->right->val)   return false;

19         queue<TreeNode*> lft;

20         lft.push(root->left);

21         queue<TreeNode*> rgt;

22         rgt.push(root->right);

23         while(( !lft.empty() )||( !rgt.empty() )){

24             int nlft=lft.size(),nrgt=rgt.size();

25             if(nlft!=nrgt)  return false;

26             for(int i=0;i<nlft;i++){

27                 TreeNode* curlft=lft.front(),*currgt=rgt.front();

28                 if( (curlft->left==NULL&&currgt->right!=NULL) ||

29                     (curlft->right==NULL&&currgt->left!=NULL) ) return false;

30                 if(curlft->left!=NULL&&

31                    currgt->right!=NULL&&

32                    curlft->left->val!=currgt->right->val)   return false;

33                 if(curlft->right!=NULL&&

34                    currgt->left!=NULL&&

35                    curlft->right->val!=currgt->left->val)   return false;

36                 if(curlft->left!=NULL)  lft.push(curlft->left);

37                 if(curlft->right!=NULL) lft.push(curlft->right);

38                 if(currgt->right!=NULL) rgt.push(currgt->right);

39                 if(currgt->left!=NULL)  rgt.push(currgt->left);

40                 lft.pop();

41                 rgt.pop();

42             }

43         }

44         return true;

45     }

46 };

47 */

48 

49 class Solution {

50 public:

51     bool isSymmetric(TreeNode *root) {

52         if(root==NULL)  return true;

53         queue<TreeNode* > lft,rgt;

54         lft.push(root->left);

55         rgt.push(root->right);

56         while((!lft.empty())||(!rgt.empty())){

57             TreeNode * curlft=lft.front(),*currgt=rgt.front();

58             lft.pop();

59             rgt.pop();

60             if(curlft==NULL&&currgt==NULL)  continue;

61             if(curlft==NULL||currgt==NULL||curlft->val!=currgt->val)  return false;

62             lft.push(curlft->left);

63             lft.push(curlft->right);

64             rgt.push(currgt->right);

65             rgt.push(currgt->left);

66         }

67         return true;

68     }

69 };

70 

71 int main()

72 {

73 

74     return 0;

75 }

View Code