Function ignoring my if condition statement - r

Question

I have a tree outlined in a data frame as:
number conc knot neg pick
1 1 0 0 1
2 1 0 0 1
3 1 0 0 1
4 3 164 0 1
5 1 0 0 1
6 1 0 0 1
7 3 159 1 1
8 0 0 0 0
9 0 0 0 0
10 3 208 1 1
11 3 181 1 1
12 3 1 1 1
13 3 95 0 1
14 0 0 0 0
15 0 0 0 0
I'm traversing the tree with a recursive function:
printtree <- function(number,tree) {
if (!is.na(tree[number,5] != 0)) {
letssee<-c(tree[number,1],tree[number,2],tree[number,3],tree[number,4],tree[number,5])
print(letssee)
}
left <- tree[number,1]
if (!is.na(left)) printtree(tree[left,1]*2,tree)
right <- tree[number,1]
if (!is.na(right)) printtree(tree[right,1]*2+1,tree)
}
My if condition should be omitting lines when the pick column = 0 but it is still printing and I can't figure out why.
Here's the output:
[1] 1 1 0 0 1
[1] 2 1 0 0 1
[1] 4 3 164 0 1
[1] 8 0 0 0 0
[1] 9 0 0 0 0
[1] 5 1 0 0 1
[1] 10 3 208 1 1
[1] 11 3 181 1 1
[1] 3 1 0 0 1
[1] 6 1 0 0 1
[1] 12 3 1 1 1
[1] 13 3 95 0 1
[1] 7 3 159 1 1
[1] 14 0 0 0 0
[1] 15 0 0 0 0
Is it ignoring my if statement because of is.na()? If I don't have the is.na check I get an error for "missing value where TRUE/FALSE needed" so it has to be there.

Answer 1

If tree[number, 5] happens to really equal zero, then the internal test
tree[number, 5] !=0
Will return FALSE. FALSE is not an NA value, so !is.na(FALSE) will always return TRUE. So the if statement as you've written it will always return TRUE if tree[number, 5] is zero.
Maybe try:
if (!is.na(X) & X !=0) {...}