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 binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isSymmetric(TreeNode *root) { if(root == NULL) return true; TreeNode *pleft = root->left; TreeNode *pright = root->right; return isMirror(pleft,pright); } private: bool isMirror(TreeNode *pleft,TreeNode *pright){ if(pleft==NULL && pright==NULL) return true; if((pleft==NULL && pright!=NULL)||(pleft!=NULL && pright==NULL)) return false; else if(pleft->val != pright->val) return false; else if(isMirror(pleft->left,pright->right)==false || isMirror(pleft->right,pright->left)==false) return false; else return true; } };
[LeetCode] Symmetric Tree,布布扣,bubuko.com
原文:http://www.cnblogs.com/Xylophone/p/3864706.html