剑指offer——面试题25：二叉树中和为某一值的路径

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); // 如果是叶结点，并且路径上结点的和等于输入的值 // 打印出这条路径 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"); } // 如果不是叶结点，则遍历它的子结点 if(pRoot->m_pLeft != NULL) FindPath(pRoot->m_pLeft, expectedSum, path, currentSum); if(pRoot->m_pRight != NULL) FindPath(pRoot->m_pRight, expectedSum, path, currentSum); // 在返回到父结点之前，在路径上删除当前结点， // 并在currentSum中减去当前结点的值 currentSum -= pRoot->m_nValue; path.pop_back(); }