面试题25：二叉树中和为某一值的路径

利用先序遍历访问树，将节点存储在栈中，没遍历一个节点比较currentSum和expectedSum，如果currentSum比expectedSum大就没必要走下去了。如果相等且是叶子节点就输出

利用栈存储路径，没有直接使用STL中的stack的原因是在stack中只能得到栈顶元素，而我们需要打印整个路径

注意形参用引用提高效率：

#include <vector>
using namespace std;

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

void FindPath(BinaryTreeNode* pRoot, int expectedSum, vector<int>& path, int& currentSum);

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, vector<int>& path, int& currentSum)
{
    if (pRoot != NULL)
    {
        currentSum += pRoot->m_nValue;
        path.push_back(pRoot->m_nValue);
    }

    if (pRoot->m_pLeft == NULL && pRoot->m_pRight == NULL && currentSum == expectedSum)
    {
        vector<int>::iterator ite = path.begin();
        while (ite != path.end())
        {
            printf("%d	", *ite);
            ite++;
        }

        printf("
");
    }

    if (currentSum < expectedSum)
    {
        if (pRoot->m_pLeft != NULL)
            FindPath(pRoot->m_pLeft, expectedSum, path, currentSum);
        if (pRoot->m_pRight != NULL)
            FindPath(pRoot->m_pRight, expectedSum, path, currentSum);
    }

    path.pop_back();
    currentSum -= pRoot->m_nValue;
}

 http://www.nowcoder.com/practice/b736e784e3e34731af99065031301bca?tpId=13&tqId=11177&rp=2&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    void find(TreeNode* root, int sum)
    {
        if (root == NULL)return;
        path.push_back(root->val);
        if (!root->left && !root->right && sum == root->val)
            res.push_back(path);
        else
        {
            if (root->left)
                find(root->left, sum - root->val);
            if (root->right)
                find(root->right, sum - root->val);
        }
        path.pop_back();
    }
    vector<vector<int>> FindPath(TreeNode* root, int expectNumber) {
        find(root, expectNumber);
        return res;
    }
};