## Amazon Interview Question for Software Engineers

Country: United States
Interview Type: In-Person

Comment hidden because of low score. Click to expand.
1
of 1 vote

``````class Node:
def __init__(self, ID, parent):
self.ID = ID
self.parentID = parent
self.left = None
self.right = None

#function to identify if given is preorder
'''
Create a stack to store nodes in the current path when traversing.
Push node[i] into stack once node[i] is verified to be valid (valid only when parent of node[i] is in stack.
In preorder a parent must show up earlier than its child)
Whenever stack top is not the parent of node[i], pop until parent of node[i] is at stack top. Push node[i].
If all nodes popped but parent of node[i] still not found, then node[i] is not in preorder sequence.
'''
def isPreorder(nodes):
if not nodes:
return True

st = [nodes[0].ID]
i = 1
while i < len(nodes):
if not st:
return False
if st[-1] is nodes[i].parentID:
st.append(nodes[i].ID)
i += 1
else:
st.pop()
return True``````

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Comment hidden because of low score. Click to expand.
0

@acoding, suppose the input is : ( parent_id, node_id) format :

``( null, 0 ), ( null, 2 ) , ( null, 4 ) ,   (0,1) , ( 2,3), ( 4,5)``

what would happen? Of course we are giving wrong input -- but..

Comment hidden because of low score. Click to expand.
1
of 1 vote

Solution in java, memory is constant and time is linear (amortized)

``````public class Solution {

// Node class
private class Node {
int ID;
Node parent;
}

public boolean isPreOrder(List<Node> nodes) {

// Assume that empty list represents correct tree
if (nodes.size() == 0) {
return true;
}

Node prev = nodes.get(0);

// Root has no parent otherwise not pre-order
if (prev.parent != null) {
return false;
}

for (int i = 1; i < nodes.size(); i++) {
Node current = nodes.get(i);

// There is only one root node
if (current.parent == null) {
return false;
}

// Find parent in current list of actively exploring
while (prev != null && current.parent.ID != prev.ID) {
prev = prev.parent;
}

// It means that we have not visited such element or we visited but we already marked subtree as done
if (prev == null) {
return false;
}

// Explore current's subtree
prev = current;
}
}

return true;
}``````

Comment hidden because of low score. Click to expand.
-1
of 1 vote

I think the keep it simple, stupid works better here.
1. Construct a graph from the list ( it is a type of adj list anyways )
2. Check if the graph is a tree first.
2.1 In that case, there must be no cycle
2.2 and one clear node with null parent.
3. Do a pre-order traversal on that "tree" and check at any point if there is any discrepancy

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