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;
};