Leetcode Same Tree

1.テーマ
Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
2.ソリューション1

class Solution {
public:
    bool isSameTree(TreeNode *p, TreeNode *q) {
        if(!p && !q)
            return true;
        else if(!p && q)
            return false;
        else if(p && !q)
            return false;
        else
        {
            if(p->val != q->val)
                return false;
            else
            {    
                queue<TreeNode*> lq;
                queue<TreeNode*> rq;
                lq.push(p);
                rq.push(q);
                while(!lq.empty() && !rq.empty())
                {
                    TreeNode* lfront = lq.front();
                    TreeNode* rfront = rq.front();
                    lq.pop();
                    rq.pop();
                    if(!lfront->left && !rfront->left)
                        ;// null
                    else if(!lfront->left && rfront->left)
                        return false;
                    else if(lfront->left && !rfront->left)
                        return false;
                    else
                    {
                        if(lfront->left->val != rfront->left->val)
                            return false;
                        else
                        {
                            lq.push(lfront->left);
                            rq.push(rfront->left);
                        }
                    }
                    if(!lfront->right && !rfront->right)
                        ;// null
                    else if(!lfront->right && rfront->right)
                        return false;
                    else if(lfront->right && !rfront->right)
                        return false;
                    else
                    {
                        if(lfront->right->val != rfront->right->val)
                            return false;
                        else
                        {
                            lq.push(lfront->right);
                            rq.push(rfront->right);
                        }
                    }    
                }
                return true;
            }
        }
    }
};

考え方:非常に簡単で、広さが遍歴すればいい.判断が煩雑で、左右のノードの値が同じでなければならない.
3.ソリューション2

class Solution {
public:
    bool isSameTree(TreeNode *p, TreeNode *q) {
        if(p == NULL && q == NULL)
		return true;
	else if(p == NULL || q == NULL)
		return false;

	bool isleftTreeSame = isSameTree(p->left, q->left);
	bool isrightTreeSame = isSameTree(p->right,q->right);

	return p->val == q->val && isleftTreeSame && isrightTreeSame;
    }
};

考え方:やはり再帰的で、遅いです.
http://www.waitingfy.com/archives/1588