leetcode中的递归调用总结

只是总结我在刷leetcode过程中遇到的使用递归来解决的问题

437. Path Sum III

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

10
/  \
5   -3
/ \    \
3   2   11
/ \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

代码：

class Solution {
public:
int pathSum(TreeNode* root, int sum) {
if(!root) return 0;
return sumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
}
private:
int sumUp(TreeNode* root, int pre, int& sum){
if(!root) return 0;
int current = pre + root->val;
return (current == sum) + sumUp(root->left, current, sum) + sumUp(root->right, current, sum);
}
};

78. Subsets

Given a set of distinct integers, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,

If nums = 

[1,2,3]

, a solution is:

[
[3],
[1],
[2],
[1,2,3],
[1,3],
[2,3],
[1,2],
[]
]

class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> subs;
vector<int> sub;
genSubsets(nums, 0, sub, subs);
return subs;
}
void genSubsets(vector<int>& nums, int start, vector<int>& sub, vector<vector<int>>& subs) {
subs.push_back(sub);
for (int i = start; i < nums.size(); i++) {
sub.push_back(nums[i]);
genSubsets(nums, i + 1, sub, subs);
sub.pop_back();
}
}
};