## Amazon Interview Question for Software Engineers

• 2

Country: United States
Interview Type: Phone Interview

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2
of 2 vote

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SOLUTION:
BFS approach does it

``````public List<Integer> leftView(TreeNode root) {
List<Integer> leftview = new ArrayList<>();
while(!q.isEmpty()) {
while(!q.isEmpty()) {
TreeNode node = q.poll();
}
q = nextLevel;
}
return leftview;
}``````

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0
of 0 vote

``````class Node():
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right

root = Node(1)
root.left = Node(2)

root.left.left = Node(4)
root.left.left.left = Node(7)
root.left.left.left.left = Node(9)

root.left.right = Node(5)
root.left.right.left = Node(8)

root.right = Node(3)
root.right.right = Node(6)

def find_left_view(rootNode):

def do_bfs(node):
p = [node]
while p:
q = []
for i, n in enumerate(p):
if i == 0:
print n.val
if n.left:
q.append(n.left)
if n.right:
q.append(n.right)
p = q

do_bfs(rootNode)

find_left_view(root)``````

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0
of 0 vote

I think DFS will work better in this scenario.

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0
of 0 vote

I think DFS will work better in this scenario.

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0
of 0 vote

Dfs

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0
of 0 vote

``````left view of the tree is nothing but the first element of in each level.
solution:
Step 1:
Print root node:
step 2:
if (root has left node)
{
print the node
push it to queue,
}
else if ( root has right node )
{
print the value
push it to queue,
}
step 3:
pop the value from queue and repeat step 2 until the queue is empty.``````

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

``````left view of the tree is nothing but the first element of in each level.
solution:
Step 1:
Print root node:
step 2:
if (root has left node)
{
print the node
push it to queue,
}
else if ( root has right node )
{
print the value
push it to queue,
}
step 3:
pop the value from queue and repeat step 2 until the queue is empty.``````

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

``````left view of the tree is nothing but the first element of in each level.
solution:
Step 1:
Print root node:
step 2:
if (root has left node)
{
print the node
push it to queue,
}
else if ( root has right node )
{
print the value
push it to queue,
}
step 3:
pop the value from queue and repeat step 2 until the queue is empty.``````

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

``````left view of the tree is nothing but the first element of in each level.
solution:
Step 1:
Print root node:
step 2:
if (root has left node)
{
print the node
push it to queue,
}
else if ( root has right node )
{
print the value
push it to queue,
}
step 3:
pop the value from queue and repeat step 2 until the queue is empty.``````

}

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

``````left view of the tree is nothing but the first element of in each level.
solution:
Step 1:
Print root node:
step 2:
if (root has left node)
{
print the node
push it to queue,
}
else if ( root has right node )
{
print the value
push it to queue,
}
step 3:
pop the value from queue and repeat step 2 until the queue is empty.``````

}

Name:

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