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();
}