# 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<>();
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) {
}

return;
}

if (node.left != null) {

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

if (node.right != null) {