C++ 4ms Two solutions: itertive level by level (with two vectors) and recursive


  • 1
    P
    bool isSymmetric(TreeNode* root) 
    {
        if(!root)
            return true;
        
        return isSymmetricRec(root->left, root->right);
    }
    
    bool isSymmetricRec(TreeNode* left, TreeNode* right) 
    {
        if(!left && !right)
            return true;
            
        if(!left || !right)
            return false;
            
        if(left->val != right->val)
            return false;
            
        bool outerPairIsSym = isSymmetricRec(left->left, right->right);
        bool innerPairIsSym = isSymmetricRec(left->right, right->left);
        return outerPairIsSym && innerPairIsSym;
    }
    
    bool isSymmetric_Iter2Vectors(TreeNode* root) 
    {
        if(!root)
            return true;
            
        vector<TreeNode*> line1, line2, * curLine = &line1, *nextLine = &line2;;
        curLine->push_back(root->left);
        curLine->push_back(root->right);
        while(!curLine->empty())
        {
            int lineLen = curLine->size();
            for(int l = 0, r = lineLen - 1; l < r; ++l, --r)
            {
                if(!(*curLine)[l] && !(*curLine)[r])
                    continue;
                if(!(*curLine)[l] || !(*curLine)[r])
                    return false;
                if((*curLine)[l]->val != (*curLine)[r]->val)
                    return false;
            }
            
            for(int i = 0; i < lineLen; ++i)
            {
                if((*curLine)[i])
                {
                    nextLine->push_back((*curLine)[i]->left);
                    nextLine->push_back((*curLine)[i]->right);    
                }
            }
            curLine->clear();
            vector<TreeNode*> * tmpCur = curLine;
            curLine = nextLine;
            nextLine = tmpCur;
        }
        return true;
    }

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