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.

 

实质是判断两颗树相同的变形，要判断一颗数是否对称，那么就是判断他的两颗子树是否对称相等，代码如下：

/**
 * 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 ) return true;
        return isSyRec(root->left, root->right);
    }
    
    bool isSyRec(TreeNode* lhs, TreeNode* rhs) {
        if( !lhs && !rhs ) return true;
        if( lhs && rhs && lhs->val == rhs->val )    //lhs左子树与rhs右子树相同，lhs右子树与rhs左子树相同
            return isSyRec(lhs->left, rhs->right) && isSyRec(lhs->right, rhs->left);
        return false;
    }
};