[Leetcode]Path Sum && Path Sum II

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 and 

sum
= 22

,

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

return true, as there exist a root-to-leaf path 

5->4->11->2

 which sum is 22.

DFS不解释

class Solution {
public:
bool hasPathSum(TreeNode *root, int sum) {
return pathSum(root,0,sum);
}
bool pathSum(TreeNode *root,int previousSum,int sum){
if(root == NULL) return false;
int currentSum = previousSum + root->val;
if(root->left == NULL && root->right == NULL){
if(currentSum == sum){
return true;
}
else{
return false;
}
}
return pathSum(root->left,currentSum,sum) || pathSum(root->right,currentSum,sum);

}
};

第二题

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]
]

还是一样的配方，还是熟悉的味道~~~~

class Solution {
public:
vector<vector<int> > result;
vector<vector<int> > pathSum(TreeNode *root, int sum) {
vector<int> onePath;
path(root, 1, onePath, sum);
return result;
}
void path(TreeNode *root, int step, vector<int> onePath, int target){
if (NULL == root) return;
if (NULL == root->left && NULL == root->right){
if (root->val == target){
onePath.push_back(root->val);
result.push_back(onePath);
}
else
return;
}
else{
onePath.push_back(root->val);
if (root->left) path(root->left, step + 1, onePath, target - root->val);
if (root->right) path(root->right, step + 1, onePath, target - root->val);
}
}
};