劍指offer——面試題25:二叉樹中和為某一值的路徑
阿新 • • 發佈:2019-01-22
void FindPath(BinaryTreeNode* pRoot, int expectedSum) { if(pRoot == NULL) return; std::vector<int> path; int currentSum = 0; FindPath(pRoot, expectedSum, path, currentSum); } void FindPath ( BinaryTreeNode* pRoot, int expectedSum, std::vector<int>& path, int& currentSum ) { currentSum += pRoot->m_nValue; path.push_back(pRoot->m_nValue); // 如果是葉結點,並且路徑上結點的和等於輸入的值 // 打印出這條路徑 bool isLeaf = pRoot->m_pLeft == NULL && pRoot->m_pRight == NULL; if(currentSum == expectedSum && isLeaf) { printf("A path is found: "); std::vector<int>::iterator iter = path.begin(); for(; iter != path.end(); ++ iter) printf("%d\t", *iter); printf("\n"); } // 如果不是葉結點,則遍歷它的子結點 if(pRoot->m_pLeft != NULL) FindPath(pRoot->m_pLeft, expectedSum, path, currentSum); if(pRoot->m_pRight != NULL) FindPath(pRoot->m_pRight, expectedSum, path, currentSum); // 在返回到父結點之前,在路徑上刪除當前結點, // 並在currentSum中減去當前結點的值 currentSum -= pRoot->m_nValue; path.pop_back(); }