一开始的思路是：中序遍历+判断遍历后的数组，时间空间都不是最优
果然超时了
1 /** 2 * Definition for binary tree 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 vector<int> inorder; 14 stack<TreeNode*> s; 15 TreeNode *cur = root; 16 while(cur != NULL || !s.empty()){ 17 while(cur != NULL){ 18 s.push(cur); 19 cur = cur->left; 20 } 21 cur = s.top(); 22 inorder.push_back(cur->val); 23 s.pop(); 24 cur = cur->right; 25 } 26 if(inorder.size()%2 != 0){ 27 int l = 0; 28 int r = inorder.size()-1; 29 while(l < r){ 30 if(inorder[l] != inorder[r]){ 31 return false; 32 } 33 } 34 return true; 35 } 36 else{ 37 return false; 38 } 39 } 40 };
这题其实考查递归的：
1 /** 2 * Definition for binary tree 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) 14 return true; 15 return Recursive(root->left,root->right); 16 } 17 bool Recursive(TreeNode *a,TreeNode *b){ 18 if(a==NULL && b==NULL)return true; 19 if(!a || !b)return false; 20 if(a->val != b->val)return false; 21 return Recursive(a->left,b->right)&&Recursive(a->right,b->left); 22 } 23 };
面试的时候如果需要改成迭代呢？
1 /** 2 * Definition for binary tree 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) 14 return true; 15 return Iterative(root); 16 } 17 bool Iterative(TreeNode *root){ 18 stack<TreeNode*> s1; 19 stack<TreeNode*> s2; 20 s1.push(root->left); 21 s2.push(root->right); 22 while(!s1.empty() && !s2.empty()){ 23 TreeNode *tmp1 = s1.top(); 24 s1.pop(); 25 TreeNode *tmp2 = s2.top(); 26 s2.pop(); 27 if((tmp1==NULL && tmp2!=NULL)||(tmp1!=NULL && tmp2==NULL)) 28 return false; 29 if(tmp1!=NULL && tmp2!=NULL){ 30 if(tmp1->val!=tmp2->val)return false; 31 s1.push(tmp1->left); 32 s1.push(tmp1->right); 33 s2.push(tmp2->right); 34 s2.push(tmp2->left); 35 } 36 } 37 return true; 38 } 39 };