Symmetric Tree

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.

思路：用递归，左边子树和右边子树相等。

看了别人代码后，才知道的。做的编程题太少了。

/** * 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 isSymmetricRecursively(TreeNode* pLeft, TreeNode* pRight){if(pLeft==NULL && pRight==NULL){return true;}if(pLeft==NULL && pRight!=NULL){return false;}if(pLeft!=NULL && pRight==NULL){return false;}if(pLeft && pRight){if(pLeft->val==pRight->val){return   isSymmetricRecursively(pLeft->left, pRight->right) && isSymmetricRecursively(pLeft->right, pRight->left);}else{return false;}}}    bool isSymmetric(TreeNode *root) {   if(root==NULL)   {   return true; //空的返回true   }   return isSymmetricRecursively(root->left, root->right);             }};