leetcode || 101、Symmetric Tree

problem：

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.

Hide Tags   Tree Depth-first Search 題意：判斷一棵二叉樹是否為鏡像二叉樹（結構對稱，數字相等）

thinking：

（1）該題合适的解法是遞歸法，采用DFS思想

遞歸，儲存左右兩個節點，然後判斷leftNode->left和rightNode->right，以及leftNode->right和rightNode->left。如此不斷遞歸

（2）還想到另外一種非遞歸法：層序周遊法，子結點為NULL時代表數值0，從左往右層序周遊得到數組a,從右往左層序周遊得到數組b

如果a==b，則該二叉樹是鏡像的。但送出時顯示：Status: 

Memory Limit Exceeded   

code：

遞歸法：

class Solution {
public:
    bool check(TreeNode *leftNode, TreeNode *rightNode)
    {
        if (leftNode == NULL && rightNode == NULL)
            return true;
            
        if (leftNode == NULL || rightNode == NULL)
            return false;
            
        return leftNode->val == rightNode->val && check(leftNode->left, rightNode->right) && 
            check(leftNode->right, rightNode->left);
    }
    
    bool isSymmetric(TreeNode *root) {
        if (root == NULL)
            return true;
            
        return check(root->left, root->right);
    }
};
           

層序周遊法： Status: 

Memory Limit Exceeded

class Solution {
private:
    vector<int> res1;
    vector<int> res2;
public:
    bool isSymmetric(TreeNode *root) {
        if(root==NULL)
            return true;
        res1.clear();
        res2.clear();
        level_walk_left(root);
        level_walk_right(root);
        if(res1==res2)
            return true;
        else
            return false;
    }
protected:
    void level_walk_left(TreeNode *root)
    {
        queue<TreeNode *> _queue;
        _queue.push(root);
        while(!_queue.empty())
        {
            TreeNode *tmp=_queue.front();
            if(tmp==NULL)
                res1.push_back(0);
            else
            {
                res1.push_back(tmp->val); 
                _queue.pop();
                if(tmp->left!=NULL)
                    _queue.push(tmp->left);
                else
                    _queue.push(NULL);
                if(tmp->right!=NULL)
                    _queue.push(tmp->right);
                else
                    _queue.push(NULL);
            }
        }
    }
    
    void level_walk_right(TreeNode *root)
    {
        queue<TreeNode *> _queue;
        _queue.push(root);
        while(!_queue.empty())
        {
            TreeNode *tmp=_queue.front();
            if(tmp==NULL)
                res2.push_back(0);
            else
            {
                res2.push_back(tmp->val); 
                _queue.pop();
                if(tmp->right!=NULL)
                    _queue.push(tmp->right);
                else
                    _queue.push(NULL);
                if(tmp->left!=NULL)
                    _queue.push(tmp->left);
                else
                    _queue.push(NULL);
            }
        }
    }
    
};