[7.4] Depth First Search(DFS) in Python | Data Structures in Python

Depth First Search algorithm is used to traverse graph or binary tree in such a way that it initially ignores the adjacent nodes and keep exploring the current node recursively.
In this tutorial we have used plain recursion to implement DFS in python.

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[Ruin.]

I've struggled with dfs and bfs and your tutorials are what I needed. Thank you

[scarytuber]

i like the way you teach, this was giving me a headache trying to figure it out until i saw your video, thank you

[smith_]

thank you so much.... for the detailed needed it the most! Keep posting such detailed explainations!

[mister_bake]

Thanks, mate. Great explanation as usual!

[arthgupta]

bro you really helped me alot keep on posting such quality and concised tutorials

[subratsingh]

precise info and good explanation ! Thanks for makign this!

[ParkootsUnplugged]

if you are not adding neighbours in graphs for D and E, it will not work when you change the root
, so you should provide all the neighbours

[devarshisingh]

could have used stack. but good explanation any way.

[muhammadiftikhar]

Dear Sir, kya ek full stack developer ko cloud ana chaheye. Or agr haan tu aws ki kon se certification krne chaheye.

[arielcototapia]

i will be very helpfully if u put any subtitles

[varunshrivastava]

In dfs does it matter from which side we start traversal left or right?

[suneosama]

E[] for neighbour in graph[E] why it work? Abecd, e doesn't have c, can anybody explain? 😅

[pcexpertsahil]

I need your help urgently ! Can you please help me with my college assignment? I have to implement breadth-first algorithm into into matrix of n by n size and value of this n is taken from the user as input. This matrix has all the values of 1's and 0's . The zeros in are randomly placed and the quantity of 0's is 10% of the total matrix size nXn.Rest All values are filled with ones. The zeroes are considered to be obstacles and job is to reach from any bottom most corner point to top most corner point preferably diagonally opposite points. I also have to do this same problem by A star search algorithm and compare the time difference between the two.