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]
]

DFS

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        if(root == null)
            return result;
        ArrayList<Integer> output = new ArrayList<Integer>();
        int curSum = 0;
        generate(curSum, sum, root, output, result);
        return result;
    }
    
    private void generate(int curSum, int target, TreeNode root, ArrayList<Integer> output, ArrayList<ArrayList<Integer>> result){
        curSum += root.val;
        output.add(root.val);
        
        Boolean isLeaf = (root.left == null && root.right == null);
        if(curSum == target && isLeaf){
            ArrayList<Integer> tmp = new ArrayList<Integer>();
            tmp.addAll(output);
            result.add(tmp);
            // 删掉一个的原因是还要去check该node的sibling 
            output.remove(output.size() - 1);
            return;
        }
        
        if(root.left != null){
            generate(curSum, target, root.left, output, result);
        }
        
        if(root.right != null){
            generate(curSum, target, root.right, output, result);
        }
        
        output.remove(output.size() - 1);
    }
}