題目：
輸入一顆二叉樹和一個整數(shù)，打印出二叉樹中結(jié)點值的和為輸入整數(shù)的所有路徑。從根結(jié)點開始往下一直到葉結(jié)點所經(jīng)過的結(jié)點形成一條路徑枷遂。
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)點之前漆枚，在路徑上刪除當前結(jié)點
currentSum -= pRoot->m_nValue;
path.pop_back();
}
