# LeetCode Notes: Binary Tree Paths

## Question

Given the `root` of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.

Example 1: Input: root = [1,2,3,null,5]

Output: ["1->2->5","1->3"]

Example 2:

Input: root = 

Output: ["1"]

Constraints:

• The number of nodes in the tree is in the range `[1, 100]`.
• `-100 <= Node.val <= 100`

## Solution

Analysis:

Depth first search, use recursion to traverse all left and right child nodes, when there are no child nodes, store the path.

Code:

``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined? 0: val)
* this.left = (left===undefined? null: left)
* this.right = (right===undefined? null: right)
*}
*/
/**
* @param {TreeNode} root
* @return {string[]}
*/
var binaryTreePaths = function(root) {
let paths = [];
const findPath = (node,path) => {

// make sure node has a value
if(!node) return

// store new node value
path += node.val

// If there is no child node, directly store the path of the current leaf node
if(node.left == null && node.right == null){
paths.push(path)
}else{
// There are child nodes, continue the depth-first search
path +='->'
findPath(node.left,path);
findPath(node.right,path)
}
}
findPath(root,'')
return paths
};
``````