[Leetcode] Path Sum II

题目：

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]
]

思路：类似path sum I，DFS，加一个vector记录经过的node，注意这里不可以剪枝。

class Solution {
public:
    int trace_sum;
    vector<int> trace;
    
    void path_sum_helper(TreeNode* root, int sum, vector<vector<int>>& result) {
        if (root == nullptr) return;
        trace_sum += root->val;
        trace.push_back(root->val);
        if (root->left == nullptr && root->right == nullptr) {
            if (trace_sum == sum) {
                result.push_back(trace);
            }
            trace_sum -= root->val;
            trace.pop_back();
            return;
        }
        path_sum_helper(root->left, sum, result);
        path_sum_helper(root->right, sum, result);
        trace_sum -= root->val;
        trace.pop_back();
    }
    
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        vector<vector<int>> result;
        path_sum_helper(root, sum, result);
        return result;
    }
};

总结：复杂度为O(2^n).