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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.

/**
 * 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;
        return isSymmetricUtil(root->left , root->right);
    }
    bool isSymmetricUtil(TreeNode *root1 , TreeNode *root2)
    {
        if(root1==NULL || root2==NULL)
            return root1==NULL && root2==NULL;
        if(root1->val == root2->val)
            return isSymmetricUtil(root1->left, root2->right) && isSymmetricUtil(root1->right , root2->left);
        else
            return false;
    }
};

//其他写法
class Solution {
public:
    bool isSymmetricUtil(TreeNode* root1, TreeNode* root2)
    {
        if(root1 == NULL && root2 == NULL) return true;
        if(root1 == NULL && root2 != NULL) return false;
        if(root1 != NULL && root2 == NULL) return false;
        return root1->val == root2->val && isSymmetricUtil(root1->left, root2->right) && isSymmetricUtil(root1->right, root2->left);
    }
    bool isSymmetric(TreeNode *root) {
        if(root == NULL) return true;
        return isSymmetricUtil(root->left, root->right);
    }
};