# Binary Tree Path Sum

Easy

## Question

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
``````

Expected result

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

## Thinking

Binary tree has three traversal methods.(refer: http://javabeat.net/binary-search-tree-traversal-java/)

• Inorder Traversal
• Preorder Traversal
• Postorder Traversal

Path definition: root to leaf
Accepted path: Summary all node values in the path equals the given number
Path: root is a leaf and root.val equals the given number
Other: set left or right node as root and target as target - root.val. call method again

Note: Don’t forget checking null node value

## Review

Using a helper method is easy to understand the code. Time complexcity is same since every node needs a copy of parents.

## Solution

#### Java (Review)

``````/**
* 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>> sumPaths = new ArrayList<List<Integer>>();
List<Integer> parents = new ArrayList<Integer>();
helper(root, target, parents, sumPaths);
return sumPaths;
}

private void helper(TreeNode node, int sum, List<Integer> parents,
List<List<Integer>> sumPaths) {
if (node == null) {
return;
}
sum = sum - node.val;
List<Integer> path = new ArrayList<Integer>(parents);
if ((node.left == null && node.right == null) && sum == 0) {
return;
}
helper(node.left, sum, path, sumPaths);
helper(node.right, sum, path, sumPaths);
}
}
``````

#### Java

``````/**
* 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>> sumPaths = new ArrayList<List<Integer>>();
if (root == null) {
return sumPaths;
} else if (root.val == target && isLeaf(root)) {
List<Integer> path = new ArrayList<Integer>();
} else {
if (root.left != null) {
// left node
List<List<Integer>> childPaths = binaryTreePathSum(root.left, target - root.val);
if (childPaths.size() > 0) {
for (List<Integer> path : childPaths) {
List<Integer> leftPath = new ArrayList<Integer>();
}
}
}
if (root.right != null) {
// right node
List<List<Integer>> childPaths = binaryTreePathSum(root.right, target - root.val);
if (childPaths.size() > 0) {
for (List<Integer> path : childPaths) {
List<Integer> rightPath = new ArrayList<Integer>();