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.
//Definition for a binary tree node. struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; //递归操作。推断一棵二叉树是否对称 class Solution { public: bool isSymmetric(TreeNode* root) { if (root == NULL) return true; else { return ifSymmetric(root->left,root->right); } } bool ifSymmetric(TreeNode* tree1, TreeNode* tree2){ if (tree1==NULL && tree2==NULL) return true; else if (tree1 == NULL || tree2 == NULL) return false; if (tree1->val != tree2->val) return false; else { return ifSymmetric(tree1->left, tree2->right) && ifSymmetric(tree1->right, tree2->left); } } };