Amazon.com Interview Question
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Interview Question for Software Development Engineer Intern at Amazon.com:
Jan 10, 2012
Print the BST in level order
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0 of 3 people found this helpful
Feb 1, 2012
Recursive function that starts at the root and recursively prints the value of each left and right node, along with the level order.
bstprint(node, count) {
if (node == null) {
return;
}
print count + ": " + node.value; //pseudocode for printing cause lazy
bstprint(node.left, count + 1);
bstprint(node.right, count + 1);
}
bstprint(root, 0); //initial call
bstprint(node, count) {
if (node == null) {
return;
}
print count + ": " + node.value; //pseudocode for printing cause lazy
bstprint(node.left, count + 1);
bstprint(node.right, count + 1);
}
bstprint(root, 0); //initial call
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Apr 9, 2012
if not tree:
return
nodes=collections.deque([tree])
currentCount, nextCount = 1, 0
while len(nodes)!=0:
currentNode=nodes.popleft()
currentCount-=1
print currentNode.val,
if currentNode.left:
nodes.append(currentNode.left)
nextCount+=1
if currentNode.right:
nodes.append(currentNode.right)
nextCount+=1
if currentCount==0:
#finished printing current level
print '\n',
currentCount, nextCount = nextCount, currentCount
return
nodes=collections.deque([tree])
currentCount, nextCount = 1, 0
while len(nodes)!=0:
currentNode=nodes.popleft()
currentCount-=1
print currentNode.val,
if currentNode.left:
nodes.append(currentNode.left)
nextCount+=1
if currentNode.right:
nodes.append(currentNode.right)
nextCount+=1
if currentCount==0:
#finished printing current level
print '\n',
currentCount, nextCount = nextCount, currentCount
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