剑指offer---二叉树中和为某一值的路径

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution {
11 public:
12     vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {
13 
14     }
15 };

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };*/
10 class Solution {
11 public:
12     vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {
13         vector<vector<int> > res;
14         vector<int> path;
15         int currentSum = 0;
16         if(root == NULL)
17             return res;
18         search(root, expectNumber, path, res, currentSum);
19         return res;
20     }
21 
22     void search(TreeNode* proot, int expectNumber, vector<int> &path,
23             vector<vector<int> > &res, int currentSum){
24 
25         currentSum += proot->val;
26         path.push_back(proot->val);
27 
28         bool isLeaf = proot->left == NULL && proot->right == NULL;
29         // 
30         if(currentSum == expectNumber && isLeaf){
31             res.push_back(path);
32             path.pop_back();
33             return ;
34         }
35         //
36         if(proot->left != NULL)
37             search(proot->left, expectNumber, path, res, currentSum);
38         if(proot->right != NULL)
39             search(proot->right, expectNumber, path, res, currentSum);
40         path.pop_back();
41     }
42 };