Path SumI、II——给出一个数，从根到子的和等于它

I、Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree andsum = 22,

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

return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

PS：要想好判断条件，递归的时候考虑到必须要到子节点。

 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     bool hasPathSum(TreeNode *root, int sum) {
13         if(root==NULL) return false;
14         return dfs(root,sum);
15     }
16     bool dfs(TreeNode *root, int sum){
17         if(root==NULL) return false;
18         if(root->left==NULL&&root->right==NULL&&sum==root->val) return true;
19         return dfs(root->left,sum-root->val)||dfs(root->right,sum-root->val);
20     }
21 };

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 andsum = 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         dfs(root,sum);
14         return res;
15     }
16     void dfs(TreeNode *root, int sum){
17         if(root==NULL) return;
18         v.push_back(root->val);
19         if(root->left==NULL&&root->right==NULL&&root->val==sum){
20             res.push_back(v);
21         }else{
22             dfs(root->left,sum-root->val);
23             dfs(root->right,sum-root->val);
24         }
25         v.pop_back();
26     }
27     vector<int> v;
28     vector<vector<int>> res;
29 };