題目：
輸入一顆二叉樹和一個(gè)整數(shù)，打印出二叉樹中結(jié)點(diǎn)值的和為輸入整數(shù)的所有路徑。從根結(jié)點(diǎn)開始往下一直到葉結(jié)點(diǎn)所經(jīng)過(guò)的結(jié)點(diǎn)形成一條路徑。

struct BinaryTreeNode {
    int                                          m_nValue;
    BinaryTreeNode*                    m_pLeft;
    BinaryTreeNode*                    m_pRight;
};

解法：

void findPath(BinaryTreeNode*  pRoot, int expectSum) {
    if (pRoot == 0)  return;
    std::vector<int>  path;
    int  currentSum = 0;  
    findPathRecursive(pRoot, expectSum, path, currentSum);
}

void findPathRecursive(BinaryTreeNode*  pRoot, int expectSum, std::vector<int>  path,  int currentSum) {
    if (pRoot == 0)  return;
    path.push_back(pRoot->m_nValue);
    currentSum += pRoot->m_nValue;
    
    bool isLeaf  =  pRoot->m_pLeft == 0 && pRoot->m_pRight == 0;
    if (expectSum == currentSum && isLeaf) {
        // find path
        print(path);
    }

    if (pRoot->m_pLeft != 0) {
        findPathRecursive(pRoot->m_pLeft, expectSum, path, currentSum);
    }
    if (pRoot->m_pRight != 0) {
        findPathRecursive(pRoot->m_pRight, expectSum, path, currentSum);
    }
    
    //返回父節(jié)點(diǎn)之前，在路徑上刪除當(dāng)前結(jié)點(diǎn)
    currentSum -= pRoot->m_nValue;
    path.pop_back();
}