Binary Tree Path Sum

LintCode-376. Binary Tree Path Sum

Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.

A valid path is from root node to any of the leaf nodes.

Example: Given a binary tree, and target = 5:

     1
    / \
   2   4
  / \
 2   3

return

[
  [1, 2, 2],
  [1, 4]
]

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root the root of binary tree
     * @param target an integer
     * @return all valid paths
     */
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        
        List<Integer> pathList = new ArrayList<>();
        pathList.add(root.val);
        pathSumHelper(root, target, result, pathList, root.val);
        
        return result;
    }
    
    void pathSumHelper(TreeNode node, int target, List<List<Integer>> result, List<Integer> pathList, int sum) {
        if (node.left == null && node.right == null) {
            if (sum == target) {
                result.add(new ArrayList<Integer>(pathList));
            }
            
            return;
        }
        
        if (node.left != null) {
            pathList.add(node.left.val);
            
            pathSumHelper(node.left, target, result, pathList, sum + node.left.val);
            pathList.remove(pathList.size() - 1);
        }
        
        if (node.right != null) {
            pathList.add(node.right.val);

            pathSumHelper(node.right, target, result, pathList, sum + node.right.val);
            pathList.remove(pathList.size() - 1);
        }
    }
    
}

Hope this helps,
Michael