（*）剑指offer 面试题25：二叉树中和为某一值的路径

题目：
输入一颗二叉树和一个整数，打印出二叉树中结点值的和为输入整数的所有路径。从根结点开始往下一直到叶结点所经过的结点形成一条路径。

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  path;
    int  currentSum = 0;  
    findPathRecursive(pRoot, expectSum, path, currentSum);
}

void findPathRecursive(BinaryTreeNode*  pRoot, int expectSum, std::vector  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);
    }
    
    //返回父节点之前，在路径上删除当前结点
    currentSum -= pRoot->m_nValue;
    path.pop_back();
}