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]

]

[Analysis]

Know your DFS!

Don’t forget to pop!

If you do pop, make sure we are always pushing one element on each recursion!

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    vector<vector > res;
    vector r;

    void pathSumHelper(TreeNode *node, int target) {
        if (!node->left && !node->right && target == node->val) {
            // node is a leaf and it's a solution
            r.push_back(node->val);
            res.push_back(r);
            return;
        }

        // internal node, also makes sure that we always push a node
        r.push_back(node->val);
        int current = target - node->val;

        if (node->right) {
            pathSumHelper(node->right, current);
            r.pop_back();
        }
        if (node->left) {
            pathSumHelper(node->left, current);
            r.pop_back();
        }
    }

public:
    vector<vector > pathSum(TreeNode *root, int sum) {
        res.clear();
        r.clear();

        if (!root)
            return res;

        pathSumHelper(root, sum);

        return res;
    }
};


		
	
			

			
							
Algorithm
backtracking
DFS
recursion