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.
分析：这道题使用递归，判断左右子结点是否相等，然后咱判断pLeft->left和pRight->right、pLeft->right和pRight->left结点是否相同。
/** * 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 check(TreeNode *pLeft,TreeNode *pRight) { if(pLeft==NULL&&pRight==NULL) return true; if(pLeft==NULL || pRight==NULL) return false; return pLeft->val==pRight->val && check(pLeft->left,pRight->right) && check(pLeft->right,pRight->left); } bool isSymmetric(TreeNode *root) { if(root==NULL) return true; return check(root->left,root->right); } };
这道题一开始使用中序遍历，然后不能Accept，出现在{1,2,3,3,#,2,#}不能过，中序遍历是有问题的，不应该这样判断。
原文：http://www.cnblogs.com/awy-blog/p/3597497.html