题目描述:

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

1
sum = 22

,

1
2
3
4
5
6
7
8
      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

return

1
2
3
4
[
   [5,4,11,2],
   [5,8,4,5]
]

与上一题类似, 只不过在DFS的过程中维护一个路径path, 保存从根节点到当前节点的值, 抵达叶子节点并且path中元素的和与sum相等时就把它加到结果中.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    vector<vector<int>> ret;
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<int> path;
        DFS(root, 0, sum, path);
        return ret;
    }
    
    void DFS(TreeNode *node, int n, int sum, vector<int> &path){
        if(!node){
            return;
        }
        path.push_back(node->val);
        if(!node->left && !node->right){
            if(n + node->val == sum){
                ret.push_back(path);
            }
        }
        else if(!node->left && node->right){
            DFS(node->right, node->val + n, sum, path);
        }
        else if(node->left && !node->right){
            DFS(node->left, node->val + n, sum, path);
        }
        else{
            DFS(node->left, node->val + n, sum, path);
            DFS(node->right, node->val + n, sum, path);
        }
        path.pop_back();
    }
};