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。

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     bool hasPathSum(TreeNode *root, int sum) {
         if(root==NULL) return false;
         return dfs(root,sum);
     }
     bool dfs(TreeNode *root, int sum){
         if(root==NULL) return false;
         if(root->left==NULL&&root->right==NULL&&sum==root->val) return true;
         return dfs(root->left,sum-root->val)||dfs(root->right,sum-root->val);
     }
 };

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]
]

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     vector<vector<int> > pathSum(TreeNode *root, int sum) {
         dfs(root,sum);
         return res;
     }
     void dfs(TreeNode *root, int sum){
         if(root==NULL) return;
         v.push_back(root->val);
         if(root->left==NULL&&root->right==NULL&&root->val==sum){
             res.push_back(v);
         }else{
             dfs(root->left,sum-root->val);
             dfs(root->right,sum-root->val);
         }
         v.pop_back();
     }
     vector<int> v;
     vector<vector<int>> res;
 };