Path Sum 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 and sum = 22,

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

return

[
   [5,4,11,2],
   [5,8,4,5]
]

Analyse: recursively do the left and right part. When a fraction meets the requirement, push it into the result.

Runtime: 12ms.

/**
 * Definition for a binary tree node.
 * 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) {
        vector<vector<int> > result;
        if(!root) return result;
        
        vector<int> temp;
        path(root, sum, result, temp);
        return result;
    }
    void path(TreeNode* root, int gap, vector<vector<int> >& result, vector<int>& temp){
        if(!root) return;
        
        temp.push_back(root->val);
        if(!root->left && !root->right){
            if(root->val == gap) result.push_back(temp);
        }
        path(root->left, gap - root->val, result, temp);
        path(root->right, gap - root->val, result, temp);
        
        temp.pop_back();
    }
};