Left View Of a Binary Tree
You have been given a binary tree of integers. You are supposed to find the left view of the binary tree. The left view of a binary tree is the set of all nodes that are visible when the binary tree is viewed from the left side.

Example:

The left view of the above binary tree is {5, 7, 14, 25}.

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow. The only line of each test case contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place. So -1 would not be a part of the tree nodes. For example, the input for the tree depicted in the below image will be:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Explanation :

Level 1 : The root node of the tree is 1 Level 2 : Left child of 1 = 2 Right child of 1 = 3 Level 3 : Left child of 2 = 4 Right child of 2 = null (-1) Left child of 3 = 5 Right child of 3 = 6 Level 4 : Left child of 4 = null (-1) Right child of 4 = 7 Left child of 5 = null (-1) Right child of 5 = null (-1) Left child of 6 = null (-1) Right child of 6 = null (-1) Level 5 : Left child of 7 = null (-1) Right child of 7 = null (-1) The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on. The input ends when all nodes at the last level are null(-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as: 1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format:

For each test case, print the left view of the given binary tree separated by a single space. Print the output of each test case in a separate line.

Note:

You don’t need to print anything; It has already been taken care of.

Constraints:

1

Question

Left View Of a Binary Tree

You have been given a binary tree of integers. You are supposed to find the left view of the binary tree. The left view of a binary tree is the set of all nodes that are visible when the binary tree is viewed from the left side.

Example:

example

The left view of the above binary tree is  {5, 7, 14, 25}.

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases follow.

The only line of each test case contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place. So -1 would not be a part of the tree nodes.

For example, the input for the tree depicted in the below image will be:

alt text

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :

Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null(-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. 
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format:

For each test case, print the left view of the given binary tree separated by a single space.

Print the output of each test case in a separate line.

Note:

You don’t need to print anything; It has already been taken care of.

Constraints:

1 <= T <= 100
0 <= N <= 3000
1 <= data <= 10^5 and data!=-1

Where ‘T’ is the number of test cases, and ‘N’ is the total number of nodes in the binary tree, and “data” is the value of the binary tree node

Time Limit: 1 sec

Accepted Answer

A level order traversal based solution can be presented here. For each level, we need to print the first node i.e. the leftmost node of that level.
Steps :
1. Make a queue of node type and push the root node in the queue.
2. While the queue is not empty, do :
2.1 Determine the current size of the queue.
2.2 Run a for loop to traverse all nodes of the current level and do :
2.2.1 Remove the topmost node from the queue.
2.2.2 If i==0 (first node of that level) , print it.
2.2.3 Push the left and right child of the node if they exist.

Time Complexity : O(n), where n is number of nodes in binary tree

Accepted Answer

Preorder Traversal Approach

The idea is to use preorder traversal and for each level of the binary tree include the leftmost node in the answer. In the preorder traversal, we will traverse the left subtree of any node before the right subtree so that all nodes to the left of the current node are visited before the nodes which are on the right of the current node. Hence for any level, we will reach the leftmost node before we reach any other node in the same level and we only need to include the leftmost node of each level in our answer.

The steps are as follows:

Let’s define a function preorderTraversal( root, level, maxLevel, leftView) for doing preorder traversal. Here “root” denotes the current node, “level” denotes the vertical distance of the node, “maxLevel” denotes the maximum level that we have encountered so far, and “leftView” is an array/list for storing the left view of the binary tree. Initialise the level of the root node to 0 and then perform the following steps:
1. If for the level of the current node, i.e.“level” is greater than the “maxLevel” then add the current node to “leftView” because this node is the leftmost node in the current level. Also, update the value of “maxLevel”.
2. Recur for the left subtree.

  preorderTraversal(root->left,level+1,maxLevel,leftView).

3. And, Recur for the right subtree.

  preorderTraversal(root->right,level+1,maxLevel,leftView).

3. At the end, “leftView” will contain all the nodes which are possible to see from the left side of the given binary tree. So return the “leftView”.

Space Complexity: O(n)Explanation:

O(N), where N is the total number of nodes in the binary tree.

We are storing the value of the leftmost nodes in the answer. So in the worst case when the binary tree is skewed, there will be a total of ‘N’ nodes to store because in that case, all the tree nodes are visible from the left side of the given binary tree. Also, the recursion stack for pre-order traversal will grow up to a maximum size of ‘N’ in the case of skewed trees. Thus, the overall space complexity is O(N).

Time Complexity: O(n)Explanation:

O(N), where N is the total number of nodes in the binary tree.

We are using the pre-order traversal in which we will visit each node of the given binary tree exactly once, and there are ‘N’ nodes to visit. Thus, the overall time complexity is O(N).

C++ (g++ 5.4)

/*

    Time Complexity - O(N)

    Space Complexity - O(N)



    Where N is the number of nodes in the Binary Tree.

*/



//    Function for doing preorder traversal of a binary tree.

void preorderTraversal(TreeNode *root, int level, int &maxLevel, vector &leftView)

{

    if (root == NULL)

    {

        return;

    }



    //    If the current level is greater than the maxlevel then store the current node.

    if (level > maxLevel)

    {

        leftView.push_back(root->data);

        maxLevel = level;

    }



    //    Recur for the left subtree.

    preorderTraversal(root->left, level + 1, maxLevel, leftView);



    //    Recur for the right subtree.

    preorderTraversal(root->right, level + 1, maxLevel, leftView);

}



vector getLeftView(TreeNode *root)

{

    

    //    For storing the left view of the binary tree.

    vector leftView;



    int maxLevel = -1;



    //    Do a preorder traversal of the binary tree.

    preorderTraversal(root, 0, maxLevel, leftView);



    //    Return the left view of the binary tree.

    return leftView;

}

Python (3.5)

'''

    Time Complexity - O(N)

    Space Complexity - O(N)



    Where N is the number of nodes in the Binary Tree.

'''



#    Function for doing preorder traversal of a binary tree.

def preorderTraversal(root, level, maxLevel, leftView):



    if (root == None):

        return maxLevel

    

    #    If the current level is greater than the maxlevel then store the current node.

    if (level > maxLevel):

    

        leftView.append(root.data)

        maxLevel = level

    

    #    Recur for the left subtree.

    maxLevel = preorderTraversal(root.left, level + 1, maxLevel, leftView)



    #    Recur for the right subtree.

    maxLevel = preorderTraversal(root.right, level + 1, maxLevel, leftView)

    

    return maxLevel





def getLeftView(root):



    #    For storing the left view of the binary tree.

    leftView = []



    maxLevel = -1



    #    Do a preorder traversal of the binary tree.

    maxlevel = preorderTraversal(root, 0, maxLevel, leftView)



    #    Return the left view of the binary tree.

    return leftView

Java (SE 1.8)

/*

    Time Complexity - O(N)

    Space Complexity - O(N)



    Where N is the number of nodes in the Binary Tree.

*/



import java.util.ArrayList;

import java.util.Queue;

import java.util.LinkedList;



public class Solution 

{

    public static ArrayList getLeftView(TreeNode root) 

    {

        

        //    For storing the left view.

        ArrayList leftView =new ArrayList();



        if (root == null) 

        {

            return leftView;

        }



        //    Queue for doing level order traversal.

        Queue> level=new LinkedList>();



        //    Push the root in the queue.

        level.add(root);

        int leftMostNode=-1;



        while (level.isEmpty() == false) 

        {

            

            //    Get the size of the current level.

            int queueSize = level.size();



            //    Traverse all nodes of the current level.

            for (int i = 0; i < queueSize; i++) 

            {

                

                //    Pop the node from the front of the queue.

                TreeNode curr = level.remove();



                //    Store the left most node of the current level.

                if (i == 0) 

                {

                    leftMostNode = curr.data;

                }



                //    Push the left child into the queue.

                if (curr.left != null) {

                    level.add(curr.left);

                }



                //    Push the right child into the queue.

                if (curr.right != null) {

                    level.add(curr.right);

                }

            }



            //    Store the left most node of the current level in the array.

            leftView.add(leftMostNode);

        }



        //    Return the left view of the given binary tree.

        return leftView;

    }

}

Accepted Answer

Level Order Traversal Approach

The idea is to do a level order traversal and for each level of the binary tree include the leftmost node in the answer. The steps are as follows:

Define a queue let’s say as “level” for doing level order traversal and an array “leftView” for storing the nodes of the left view of the given binary tree.
Push the root of the given tree into the ”level”.
We will keep doing the following operations until “level” does not become empty:
1. Get the size of the queue, i.e. the total number of nodes at the current level. Also declare a variable “leftMostNode” to store the left most node of the current level.
2. Visit all the nodes one by one which are at the current level and store the value of the first node(i.e. the leftmost node) of the current level in “leftMostNode”. Push their left and right child into the queue if they exist.
3. Insert the value of “leftMostNode” in “leftView”.
In the end, “leftView” will contain all the nodes which are possible to see from the left side of the given binary tree. So return the “leftView”.

Space Complexity: O(n)Explanation:

O(N), where N is the total number of nodes in the binary tree.

We are storing the value of the leftmost nodes in the answer. So in the worst case when the binary tree is skewed, there will be a total of ‘N’ nodes to store because in that case, all the tree nodes are visible from the left side of the given binary tree. Also, for doing level order traversal, we are using a queue that can have a maximum of ‘N/2’ nodes in the case of a complete binary tree. Thus, the total overall space complexity is O(N).

Time Complexity: O(n)Explanation:

O(N), where N is the total number of nodes in the binary tree.

We are using the level order traversal, in which there will be ‘N’ push and ‘N’ pop operations, where push and pop operation will take constant time for the queue data structure. Thus, the total overall time complexity will be O(N).

Python (3.5)

'''

	Time Complexity - O(N)

	Space Complexity - O(N)



	Where N is the number of nodes in the Binary Tree.

'''





# Binary tree node class for reference

class BinaryTreeNode:

    def __init__(self, data):

        self.data = data

        self.left = None

        self.right = None





def getLeftView(root):



	#    For storing the left view.

	leftView = []



	if (root == None):

		return leftView

	

	#    Queue for doing level order traversal.

	level = []



	#    append the root in the queue.

	level.append(root)

	leftMostNode = None



	while (len(level) != 0):

	

		#    Get the size of the current level.

		queueSize = len(level)



		#    Traverse all nodes of the current level.

		for i in range(queueSize):

		

			#    Pop the node from the front of the queue.

			curr = level[0]

			level.pop(0)



			#    Store the left most node of the current level.

			if (i == 0):

				leftMostNode = curr.data



			#    append the left child into the queue.

			if (curr.left != None):

				level.append(curr.left)



			#    append the right child into the queue.

			if (curr.right != None):

				level.append(curr.right)



		#    Store the left most node of the current level in the array.

		leftView.append(leftMostNode)

	

	#    Return the left view of the given binary tree.

	return leftView

C++ (g++ 5.4)

/*

    Time Complexity - O(N)

    Space Complexity - O(N)



    Where N is the number of nodes in the Binary Tree.

*/



vector getLeftView(TreeNode *root)

{

    

    //    For storing the left view.

    vector leftView;



    if (root == NULL)

    {

        return leftView;

    }



    //    Queue for doing level order traversal.

    queue *> level;



    //    Push the root in the queue.

    level.push(root);

    int leftMostNode;



    while (level.empty() == false)

    {

        

        //    Get the size of the current level.

        int queueSize = level.size();



        //    Traverse all nodes of the current level.

        for (int i = 0; i < queueSize; i++)

        {

            

            //    Pop the node from the front of the queue.

            TreeNode *curr = level.front();

            level.pop();



            //    Store the left most node of the current level.

            if (i == 0)

            {

                leftMostNode = curr->data;

            }



            //    Push the left child into the queue.

            if (curr->left != NULL)

            {

                level.push(curr->left);

            }



            //    Push the right child into the queue.

            if (curr->right != NULL)

            {

                level.push(curr->right);

            }

        }



        //    Store the left most node of the current level in the array.

        leftView.push_back(leftMostNode);

    }



    //    Return the left view of the given binary tree.

    return leftView;

}

Java (SE 1.8)

/*

    Time Complexity - O(N)

    Space Complexity - O(N)



    Where N is the number of nodes in the Binary Tree.

*/



import java.util.ArrayList;

import java.util.Queue;

import java.util.LinkedList;



public class Solution 

{

    public static ArrayList getLeftView(TreeNode root) 

    {

        

        //    For storing the left view.

        ArrayList leftView =new ArrayList();



        if (root == null) 

        {

            return leftView;

        }



        //    Queue for doing level order traversal.

        Queue> level=new LinkedList>();



        //    Push the root in the queue.

        level.add(root);

        int leftMostNode=-1;



        while (level.isEmpty() == false) 

        {

            

            //    Get the size of the current level.

            int queueSize = level.size();



            //    Traverse all nodes of the current level.

            for (int i = 0; i < queueSize; i++) 

            {

                

                //    Pop the node from the front of the queue.

                TreeNode curr = level.remove();



                //    Store the left most node of the current level.

                if (i == 0) 

                {

                    leftMostNode = curr.data;

                }



                //    Push the left child into the queue.

                if (curr.left != null) {

                    level.add(curr.left);

                }



                //    Push the right child into the queue.

                if (curr.right != null) {

                    level.add(curr.right);

                }

            }



            //    Store the left most node of the current level in the array.

            leftView.add(leftMostNode);

        }



        //    Return the left view of the given binary tree.

        return leftView;

    }

}

You have been given a binary tree of integers. You are supposed to find the left view of the binary tree. The left view of a binary tree is the set of all nodes that are visible when the binary tree is viewed from the left side.

Example:

Input Format:

Explanation :

Note :

Output Format:

Note:

Constraints:

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