# Binary Tree Paths

Easy

## Question

Given a binary tree, return all root-to-leaf paths.

Example
Given the following binary tree:

``````   1
/   \
2     3
\
5
``````

All root-to-leaf paths are:

``````[
"1->2->5",
"1->3"
]
``````

## Review

My Original implementation is using recursive in the methods. It needs more time since every node goes through all children’s paths. The review version is using a helper method and pass parent path to children.

## Solution

#### Java (review version)

``````/**
* 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 the binary tree
* @return all root-to-leaf paths
*/
public List<String> binaryTreePaths(TreeNode root) {
List<String> paths = new ArrayList<String>();
helper(root, "", paths);
return paths;
}

private void helper(TreeNode node, String parentPath, List<String> paths) {
if (node == null) {
return;
}
String path = parentPath + node.val;
if (node.left == null && node.right == null) {
}
helper(node.left, path + "->", paths);
helper(node.right, path + "->", paths);
return;
}
}
``````

#### 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 the binary tree
* @return all root-to-leaf paths
*/
public List<String> binaryTreePaths(TreeNode root) {
List<String> paths = new ArrayList<String>();
if (root == null) {
return paths;
}
if (root.left == null && root.right == null) {