打印二叉树中和为某一值的路径

输入一个二叉树，查找该树的所有路径（从根结点到叶结点的通路），并返回和（路径上所有结点值的和）为某一指定值的路径。

 1 /////////////二叉树中和为某一值的路径/////////////////////

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

 3 {

 4     if (pRoot == NULL)

 5     {

 6         return;

 7     }

 8     currentSum += pRoot->m_nValue;

 9     path.push_back(pRoot->m_nValue);

10     //如果是叶子结点，且结点的和等于希望的值，打印出这条路径

11     if (currentSum == expectedSum && pRoot->m_pLeft == NULL && pRoot->m_pRight == NULL)

12     {

13         cout<<"Find a path : ";

14         vector<int>::iterator iter = path.begin();

15         for (;iter != path.end() ; iter++)

16         {

17             cout<<*iter<<" ";

18         }

19         cout<<endl;

20     }

21     if (pRoot->m_pLeft)

22     {

23         FindPath(pRoot->m_pLeft ,expectedSum ,path ,currentSum);

24     }

25     if (pRoot->m_pRight)

26     {

27         FindPath(pRoot->m_pRight ,expectedSum ,path ,currentSum);

28     }

29     //currentSum = currentSum - path.back();

30     path.pop_back();

31 }

32 void FindPath(BinaryTreeNode* pRoot , int expectedSum)//用户接口

33 {

34     if (pRoot == NULL)

35     {

36         return;

37     }

38     int currentSum = 0 ;

39     vector<int> vec ;

40     FindPath(pRoot ,expectedSum ,vec , currentSum );

41

42 }