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Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
For example:sum = 22
,
5 / 4 8 / / 11 13 4 / \ / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int> > pathSum(TreeNode *root, int sum) { 13 vector<int> tmp; 14 vector<vector<int> > result; 15 if(root==NULL) 16 { 17 return result; 18 } 19 DFS(root,sum,tmp,result); 20 return result; 21 22 } 23 24 void DFS(TreeNode *root,int &sum, vector<int> tmp,vector<vector<int> > &result,int cur=0) 25 { 26 if(root==NULL) 27 { 28 return; 29 } 30 31 int val=root->val; 32 tmp.push_back(val); 33 34 if(root->left==NULL&&root->right==NULL) 35 { 36 if(cur+val==sum) result.push_back(tmp); 37 return; 38 } 39 DFS(root->left,sum,tmp,result,cur+val); 40 DFS(root->right,sum,tmp,result,cur+val); 41 } 42 };
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原文地址:http://www.cnblogs.com/reachteam/p/4199319.html