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.
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 };
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 }