leetcode-9

Subsets

Bit manipulation and map can be useful aside from backtracking

vector> subsets(vector& nums) {
        vector> results(pow(2,nums.size()), vector());
        for(int i=1;i<=pow(2,nums.size());i++){
            for(int j=0;j>(nums.size()-1))==1){
                    results[i-1].push_back(nums[j]);
                }
            }
        }
        return results;
    }

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Path Sum

Don't suppose that all elements are positive.
Pay attention to problem saying "root to leaf"

Path Sum III

class Solution {
public:
    int help(TreeNode* root, int sum, unordered_map& store, int pre) {
        if (!root) return 0;
        root->val += pre;
        int res = (root->val == sum) + (store.count(root->val - sum) ? store[root->val - sum] : 0);
        store[root->val]++;
        res += help(root->left, sum, store, root->val) + help(root->right, sum, store, root->val);
        store[root->val]--;
        return res;
    }

    int pathSum(TreeNode* root, int sum) {
        unordered_map store;
        return help(root, sum, store, 0);
    }
};

Up-to-bottom is faster and smaller than Bottom-to-Up as no need to store whole sums only need to store current value needed.
Use hashmap to store value makes searching occurrence faster.