非递归：层序遍历，然后为NULL叶节点填一个不可能的值用来判断；
1 /** 2 * Definition for a binary tree node. 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 queue<TreeNode*> q; 15 q.push(root); 16 while(!q.empty()){ 17 vector<int> ivec; 18 int len=q.size(); 19 for(int i=0;i<len;i++){ 20 TreeNode* front=q.front(); 21 q.pop(); 22 if(front->left){ 23 q.push(front->left); 24 ivec.push_back(front->left->val); 25 }else{ 26 ivec.push_back(-1000000); 27 } 28 if(front->right){ 29 q.push(front->right); 30 ivec.push_back(front->right->val); 31 }else{ 32 ivec.push_back(-1000000); 33 } 34 } 35 for(int i=0;i<ivec.size()-1-i;i++){ 36 int j=ivec.size()-1-i; 37 if(ivec[i]!=ivec[j]){ 38 return false; 39 } 40 } 41 } 42 return true; 43 } 44 };
递归方法：使用递归中序遍历，对NULL叶节点填充不可能数值，后进行对称判断；
1 /** 2 * Definition for a binary tree node. 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 vector<int> ivec; 15 dfs(root,ivec); 16 for(int i=0;i<ivec.size()-1-i;i++){ 17 int j=ivec.size()-1-i; 18 if(ivec[i]!=ivec[j]) return false; 19 } 20 return true; 21 } 22 void dfs(TreeNode* root,vector<int>& ivec){ 23 if(root==NULL) return; 24 dfs(root->left,ivec); 25 if(root->left) 26 ivec.push_back(root->left->val); 27 else 28 ivec.push_back(-1000000); 29 if(root->right) 30 ivec.push_back(root->right->val); 31 else 32 ivec.push_back(-1000000); 33 dfs(root->right,ivec); 34 } 35 };