113. 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]
]

Solution:

Use DFS to find the leaf of the tree.

If the sum of all nodes are the sum we want, we add this list to the result.



Code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        List<Integer> list = new ArrayList<>();
        helper(root, sum, result, list);
        return result;
    }
    
    public void helper(TreeNode root, int target, List<List<Integer>> result, List<Integer> list) {
        
        if (root == null) {
            return;
        }
        
        list.add(root.val);
        
        if (root.left == null && root.right == null && root.val == target) {
            result.add(new ArrayList<Integer>(list));
        }
 
        helper(root.left, target - root.val, result, list);
        helper(root.right, target - root.val, result, list);
        
        list.remove(list.size() - 1);
    }
}