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.
两个函数,一个是判断相等,一个是交换左右子树
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isEqual(TreeNode *root1, TreeNode *root2) { if(!root1&&!root2) return true; if(!root1||!root2) return false; if(!root1->left&&!root1->right&&!root2->left&&!root2->right&&root1->val==root2->val) return true; if(isEqual(root1->left, root2->left)&&isEqual(root1->right, root2->right)&&root1->val==root2->val) return true; return false; } void swapLeftRight(TreeNode *root) { if(NULL==root||(NULL==root->left&&NULL==root->right))return; swapLeftRight(root->left); swapLeftRight(root->right); TreeNode *temp = root->left; root->left = root->right; root->right = temp; } bool isSymmetric(TreeNode *root) { if(!root) return true; swapLeftRight(root->left); if(isEqual(root->left,root->right)) return true; return false; } };