Path Sum II

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]
]

分析：此题可以用DFS解，当遍历到叶节点时，判断当前path的sum是否等于target sum，如果相等便将当前path加入到result中。
代码如下：
 1 class Solution {
 2 public:
 3     vector<vector<int> > pathSum(TreeNode *root, int sum) {
 4         vector<vector<int>> res;
 5         if(!root) return res;
 6         vector<int> path;
 7         dfs(res,path,root,0,sum);
 8         return res;
 9     }
10     void dfs(vector<vector<int>> & res, vector<int> & path,TreeNode * root, int pre_sum, int sum){
11         if(!root) return;
12         path.push_back(root->val);
13         int cur_sum = pre_sum + root->val;
14         dfs(res,path,root->left,cur_sum,sum);
15         dfs(res,path,root->right,cur_sum,sum);
16         if(cur_sum == sum && !root->left && !root->right) res.push_back(path);
17         path.pop_back();
18     }
19 };

 

posted on 2014-08-13 13:30  Ryan-Xing  阅读(117)  评论(0)  编辑  收藏  举报