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SOLUTION:
BFS approach does it

public List<Integer> leftView(TreeNode root) {
        List<Integer> leftview = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        if(root != null) q.add(root);
        while(!q.isEmpty()) {
            leftview.add(q.peek().val);
            Queue<TreeNode> nextLevel = new LinkedList<>();
            while(!q.isEmpty()) {
                TreeNode node = q.poll();
                if(node.left != null) nextLevel.add(node.left);
                if(node.right != null) nextLevel.add(node.right);
            }
            q = nextLevel;
        }
        return leftview;
    }

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)

I think DFS will work better in this scenario.

I think DFS will work better in this scenario.

Dfs

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.

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.

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.

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.

}

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.

}