Question

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

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]
]

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<List<Integer>> rlist = new ArrayList<>();

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        
        treePath(root,new ArrayList(),sum);
        return rlist;
    }
    
    private void treePath(TreeNode root,List list,int sum){
        if(root == null) return;
        if(root.left == null && root.right == null){
            if(sum - root.val == 0){
              List<Integer> copy = new LinkedList<>(list);
              copy.add(root.val);
              rlist.add(copy);
            }
            return;
        }
        
         list.add(root.val);
        
         treePath(root.left,list,sum - root.val);
         treePath(root.right,list,sum - root.val);
         list.remove(list.size()-1);        
    }
}