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.
Analyse: the same as Same Tree
1. Recursion
Runtime: 8ms.
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) return true; 14 return symmetric(root->left, root->right); 15 } 16 bool symmetric(TreeNode* leftNode, TreeNode* rightNode){ 17 if(!leftNode && !rightNode) return true; 18 if(!leftNode || !rightNode) return false; 19 20 return leftNode->val == rightNode->val && 21 symmetric(leftNode->left, rightNode->right) && 22 symmetric(leftNode->right, rightNode->left); 23 } 24 };
2. Iteration
Runtime: 4ms.
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) return true; 14 stack<TreeNode* > stk; 15 stk.push(root->left); 16 stk.push(root->right); 17 18 while(!stk.empty()){ 19 TreeNode* node1 = stk.top(); 20 stk.pop(); 21 TreeNode* node2 = stk.top(); 22 stk.pop(); 23 24 if(!node1 && !node2) continue; 25 if(!node1 || !node2) return false; 26 if(node1->val != node2->val) return false; 27 28 stk.push(node1->left); 29 stk.push(node2->right); 30 stk.push(node1->right); 31 stk.push(node2->left); 32 } 33 return true; 34 } 35 };