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

 

思路：判断树是否是对称的 递归判断即可

class Solution {public:    bool isSymmetric(TreeNode* root) {        if(NULL == root)   return true;        TreeNode * L = root->left;        TreeNode * R = root->right;        return isSymmetricPart(L, R);    }     bool isSymmetricPart(TreeNode* L, TreeNode* R)     {         if(NULL == L && NULL == R) return true;         if(NULL == L && NULL != R || NULL == R && NULL != L) return false;         if(L->val == R->val)             return isSymmetricPart(L->left, R->right) && isSymmetricPart(L->right, R->left);         else             return false;              }};

 

更简短的写法

class Solution {public:    bool isSymmetric(TreeNode* root) {        return !root ? true : isSymmetricHelper(root->left, root->right);    }    bool isSymmetricHelper(TreeNode* left, TreeNode* right) {        if (!left && !right) { return true; }        return             (left && right) &&            (left->val == right->val) &&            isSymmetricHelper(left->left, right->right) &&            isSymmetricHelper(left->right, right->left);    }};