剑指offer——面试题25：二叉树中和为某一值的路径

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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();}

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