Please help dfs function is returning in reverse order#dfs$($a$)$ performs a depth first search starting at node $0$ and returns a list of nodes in the order in which they were seen, with start and stop times. Press shift enter to test your code. Ensure that your code has been saved first by pressing shift$+$enter on the previous cell.

from IPython.core.display import display, HTML

def dfs$\_$test$()$:test$\_$cases $=[([[1,3],[0],[1,3],[2]],[(0,(1,8)),(1,(2,3)),(3,(4,7)),(2,(5,6))]),([[],[0,3],[1],[]],[(0,(1,2)),(1,(4,7)),(3,(5,6)),(2,(9,10))]),$for $($test$\_$graph, solution$)$ in test$\_$cases: if $($solution $!=$ output$)$: s$2=\text{'}\text{'}+$ str$($solution$)+\text{'}$ Your code output: $\text{'}+$ str$($output$)+""$ failed $=$ True display$($HTML$(\text{'}\text{'}))$ display$($HTML$(\text{'}\text{'}))$

dfs$\_$test$()/$tmp$/$ipykernel$\_124/474394405.$py:$3$: DeprecationWarning: Importing display from IPython.core.display is deprecated since IPython $7\mathrm{.}14,$ please import from IPython display

from IPython.core.display import display, HTML

Failed $-$ test case: Inputs: graph $=\left[\begin{array}{ccc}1& 2& 3\\ 0& 2& 3\\ 0& 1& 3\\ 0& 1& 2\end{array}\right]$

Expected Output: $(0,(1,8)),(1,(2,7)),(2,(3,6)),(3,(4,5))$ Your code output: $(3,(4,5)),(2,(3,6)),(1,(2,7)),(0,(1,8))$

Failed $-$ test case: Inputs: graph $=\left[\begin{array}{cc}1& 3\\ 0& ?\\ 1& 3\\ 2& ?\end{array}\right]$

Expected Output: $(0,(1,8)),(1,(2,3)),(3,(4,7)),(2,(5,6))$ Your code output: $(1,(2,3)),(2,(5,6)),(3,(4,7)),(0,(1,8))$

Failed $-$ test case: Inputs: graph $=\left[\begin{array}{cc}?& ?\\ 0& 2\\ 3& ?\\ 1& ?\end{array}\right]$

Expected Output: $(0,(1,2)),(1,(4,9)),(2,(5,8)),(3,(6,7))$ Your code output: $(0,(1,2)),(3,3,4)$

Failed $-$ test case: Inputs: graph $=[],[0,3],[1],[]]$

Expected Output: $(0,(1,2)),(1,(4,7)),(3,(5,6)),(2,(9,10))$ Your code output: $(0,(1,2)),(3,(2,3)),(1,(1,4)),(2,(1,2))$

Failed $-$ test case: Inputs: graph $=[[1,2],[4,5],[3,4],[8,9],[7,8],[6,7],[,[],[,[]]$

#INPUT: $[[1],[2],[0]]($a $3$ node cycle$)$

#OUTPUT: $[(0,(1,6)),(1,(2,5)),(2,(3,4))]$

def dfs$\_$visit$($a$,$ u$,$ color, d$,$ f$,$ time, result$)$:d$[$u$]=$ time$[0]$for v in a$[$u$]$: dfs$\_$visit$($a$,$ v$,$ color, d$,$ f$,$ time, result$)$time$[0]+=1$result.append$(($u$,($d$[$u$],$ f$[$u$])))$

def dfs$($a$)$:color $=[\text{'}$WHITE$\text{'}]*$ nf $=[0]*$ nfor u in range$($n$)$: dfs$\_$visit$($a$,$ u$,$ color, d$,$ f$,[0],$ result$)$# result.sort$($key$=$lambda x: x$[1][0])$