24、剑指offer--二叉树中和为某一值的路径

 1 /*struct TreeNode {
 2     int val;
 3     struct TreeNode *left;
 4     struct TreeNode *right;
 5     TreeNode(int x) :
 6             val(x), left(NULL), right(NULL) {
 7     }
 8 };*/
 9 class Solution {
10 public:
11     vector<vector<int> > FindPath(TreeNode* root,int expectNumber) {
12         vector<int> tmp;
13         vector<vector<int> >result;
14         if(root == NULL || expectNumber<=0)
15             return result;
16         IsPath(root,expectNumber,result,tmp);
17         return result;
18     }
19     void IsPath(TreeNode* root,int expectNumber,vector<vector<int> > &findPath,vector<int> &path)//不是取地址的话，结果不正确
20     {
21         path.push_back(root->val);
22         if(root == NULL)
23             return;
24         bool isLeaf = (root->left == NULL)&&(root->right == NULL);
25         if((root->val == expectNumber)&&isLeaf)
26         {
27             findPath.push_back(path);
28         }
29         else
30         {
31             if(root->left != NULL) IsPath(root->left,expectNumber-root->val,findPath,path);
32             if(root->right != NULL) IsPath(root->right,expectNumber-root->val,findPath,path);
33         }
34         path.pop_back();
35     }
36 };