---
id: 587d825d367417b2b2512c96
title: Depth-First Search
challengeType: 1
---

## Description
<section id='description'>
Similar to <dfn>breadth-first search</dfn>, here we will learn about another graph traversal algorithm called <dfn>depth-first search</dfn>.
Whereas the breadth-first search searches incremental edge lengths away from the source node, <dfn>depth-first search</dfn> first goes down a path of edges as far as it can.
Once it reaches one end of a path, the search will backtrack to the last node with an un-visited edge path and continue searching.
Visually, this is what the algorithm is doing where the top node is the starting point of the search.
<img class='img-responsive' src='https://camo.githubusercontent.com/aaad9e39961daf34d967c616edeb50abf3bf1235/68747470733a2f2f75706c6f61642e77696b696d656469612e6f72672f77696b6970656469612f636f6d6d6f6e732f372f37662f44657074682d46697273742d5365617263682e676966'>
A simple output of this algorithm is a list of nodes which are reachable from a given node. So when implementing this algorithm, you'll need to keep track of the nodes you visit.
</section>

## Instructions
<section id='instructions'>
Write a function <code>dfs()</code> that takes an undirected, adjacency matrix <code>graph</code> and a node label <code>root</code> as parameters. The node label will just be the numeric value of the node between <code>0</code> and <code>n - 1</code>, where <code>n</code> is the total number of nodes in the graph.
Your function should output an array of all nodes reachable from <code>root</code>.
</section>

## Tests
<section id='tests'>

```yml
tests:
  - text: The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>1</code> should return an array with <code>0</code>, <code>1</code>, <code>2</code>, and <code>3</code>.
    testString: assert.sameMembers((function() { var graph = [[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]; return dfs(graph, 1);})(), [0, 1, 2, 3], 'The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>1</code> should return an array with <code>0</code>, <code>1</code>, <code>2</code>, and <code>3</code>.');
  - text: The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>1</code> should return an array with four elements.
    testString: assert((function() { var graph = [[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]; return dfs(graph, 1);})().length === 4, 'The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>1</code> should return an array with four elements.');
  - text: The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]</code> with a start node of <code>3</code> should return an array with <code>3</code>.
    testString: assert.sameMembers((function() { var graph = [[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]; return dfs(graph, 3);})(), [3], 'The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]</code> with a start node of <code>3</code> should return an array with <code>3</code>.');
  - text: The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]</code> with a start node of <code>3</code> should return an array with one element.
    testString: assert((function() { var graph = [[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]; return dfs(graph, 3);})().length === 1, 'The input graph <code>[[0, 1, 0, 0], [1, 0, 1, 0], [0, 1, 0, 0], [0, 0, 0, 0]]</code> with a start node of <code>3</code> should return an array with one element.');
  - text: The input graph <code>[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>3</code> should return an array with <code>2</code> and <code>3</code>.
    testString: assert.sameMembers((function() { var graph = [[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]; return dfs(graph, 3);})(), [2, 3], 'The input graph <code>[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>3</code> should return an array with <code>2</code> and <code>3</code>.');
  - text: The input graph <code>[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>3</code> should return an array with two elements.
    testString: assert((function() { var graph = [[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]; return dfs(graph, 3);})().length === 2, 'The input graph <code>[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>3</code> should return an array with two elements.');
  - text: The input graph <code>[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>0</code> should return an array with <code>0</code> and <code>1</code>.
    testString: assert.sameMembers((function() { var graph = [[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]; return dfs(graph, 0);})(), [0, 1], 'The input graph <code>[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>0</code> should return an array with <code>0</code> and <code>1</code>.');
  - text: The input graph <code>[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>0</code> should return an array with two elements.
    testString: assert((function() { var graph = [[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]; return dfs(graph, 0);})().length === 2, 'The input graph <code>[[0, 1, 0, 0], [1, 0, 0, 0], [0, 0, 0, 1], [0, 0, 1, 0]]</code> with a start node of <code>0</code> should return an array with two elements.');

```

</section>

## Challenge Seed
<section id='challengeSeed'>

<div id='js-seed'>

```js
function dfs(graph, root) {

}

var exDFSGraph = [
  [0, 1, 0, 0],
  [1, 0, 1, 0],
  [0, 1, 0, 1],
  [0, 0, 1, 0]
];
console.log(dfs(exDFSGraph, 3));
```

</div>



</section>

## Solution
<section id='solution'>


```js
function dfs(graph, root) { var stack = []; var tempV; var visited = []; var tempVNeighbors = []; stack.push(root); while (stack.length > 0) { tempV = stack.pop(); if (visited.indexOf(tempV) == -1) { visited.push(tempV); tempVNeighbors = graph[tempV]; for (var i = 0; i < tempVNeighbors.length; i++) { if (tempVNeighbors[i] == 1) { stack.push(i); }}}} return visited;}
```

</section>