【LeetCode】Symmetric Tree

description:
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.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as  "{1,2,3,#,#,4,#,#,5}"
code : (build a symmetric tree first, then check if the two trees are the same )

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(root == NULL) 
            return true;
        TreeNode *syroot = new TreeNode(root->val);
        build(syroot->left,root->right);
        build(syroot->right,root->left);
        if(isSameTree(syroot,root))
            return true;
        else return false;
        
    }
    void build(TreeNode *&syroot,TreeNode *root)
    {
        if(root == NULL)
        {
            syroot = NULL;
            return;
        }
        syroot = new TreeNode(root->val);
        build(syroot->left,root->right);
        build(syroot->right,root->left);
        
    }
    bool isSameTree(TreeNode *p, TreeNode *q) 
    {  
        // Start typing your C/C++ solution below  
        // DO NOT write int main() function  
        if(p == NULL && q == NULL) return true;  
        if((p == NULL && q !=NULL) || (p !=NULL && q == NULL)) return false;  
        if(p->val != q->val) return false;  
        if(!issame(p->left,q->left)) return false;  
        if(!issame(p->right,q->right)) return false;  
        return true;  
          
    }  
    bool issame(TreeNode *p, TreeNode *q)  
    {  
        if((p == NULL && q !=NULL) || (p !=NULL && q == NULL)) return false;  
        if(p == NULL && q == NULL) return true;  
        if(p->val != q->val) return false;  
        if(!issame(p->left,q->left)) return false;  
        if(!issame(p->right,q->right)) return false;  
        return true;  
          
    } 
};

아래의 코드는 더욱 간결하여 대칭 트리를 만들지 않아도 된다. 그러나 시간의 복잡도는 그다지 최적화되지 않았다

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(root == NULL) 
            return true;
        return isSame(root,root);
        
    }
    bool isSame(TreeNode *syroot,TreeNode *root)
    {
        if(syroot == NULL && root == NULL)
            return true;
        else if(syroot == NULL || root == NULL)
            return false;
        if(syroot->val != root->val)
            return false;
        return isSame(syroot->left,root->right) && isSame(syroot->right,root->left);
    }
     
};