Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and sum = 22
,
5 / 4 8 / / 11 13 4 / / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
思考:DFS。
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { private: vector<vector<int> > ret; public: void DFS(TreeNode *root,int sum,int total,vector<int> &ans) { if(root) { total+=root->val; ans.push_back(root->val); if(total==sum&&!root->left&&!root->right) { ret.push_back(ans); } DFS(root->left,sum,total,ans); ans.pop_back(); DFS(root->right,sum,total,ans); ans.pop_back(); } else ans.push_back(0); } vector<vector<int> > pathSum(TreeNode *root, int sum) { ret.clear(); vector<int> ans; DFS(root,sum,0,ans); return ret; } };