I have understood the approach but

#1

i got the approach but is there any other way to solve the problem instead of using maps?

#2

@Manvik-Arya-4391661007540658 yes it can be done without maps also,
maintain two arrays A,B
let current horizontal point(hp) be 0 for root
after that whenever you go right, do hp+1 and for left do hp-1,
obviously if you are somewhere left, hp is -ve and on right hp is +ve,
now if hp is -ve and abs(hp)>size(A) then insert it,
else ignore as a node already exists on top of it!
similar do with hp is +ve and abs(hp)>size(B),
answer will be (reverse A) + (root->data) + (B)
complexity is O(n)// both space and time

#3

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