Last Updated: 7 Oct, 2020

Maximum Width In Binary Tree

Moderate

Asked in companies

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Problem statement

You have been given a Binary Tree of integers. You are supposed to return the maximum width of the given Binary Tree. The maximum width of the tree is the maximum width among all the levels of the given tree.

The width of one level is defined as the length between the leftmost and the rightmost, non-null nodes in the level, where the null nodes in between the leftmost and rightmost are excluded into length calculation.

For example :

For the given binary tree

Example

The maximum width will be at the third level with the length of 3, i.e. {4, 5, 6}.

Input Format :

The only line of input 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.

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

Example

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 :

Print a single line that contains a single integer that denotes the maximum width for the given tree.

Note :

You do not need to print anything; it has already been taken care of. Just implement the given function.

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Approaches

01 Approach

The straightforward intuition is that first, find the maximum levels possible in the given tree, called the "HEIGHT" of the tree. And then, for each level in height, we will find the number of nodes in each level. And the maximum number of the node among all levels will be the maximum width of the given binary tree. So the implementation of our intuition takes bellow steps:

First, find the height of the tree. And for finding the height of the tree, steps are:

If the root is "NULL", return 0 as height is 0.
Else, call the recursive function for left subtree and right subtree and which has a maximum height, which would be the height of the subtree for the "ROOT" node. So the height of the "ROOT" node will be
- 1+max("LEFT-HEIGHT", “RIGHT-HEIGHT”)
Return the height of the "ROOT" node.

2. We will go for each level in [1: "HEIGHT"] and find how many nodes it contains at that level. The maximum number of nodes among all levels will be the maximum width of the given tree. For finding the number of nodes at any level, let's say "LEVEL", Steps are as follows:

If the root is "NULL", return 0 as there will be no node possible.
If "level" becomes 1, then return 1 as this root node is a node in required level.
Else, call recursive on left subtree with "level-1" and store in "LEFTNODES" and call recursive on right subtree with "level-1" and store in "RIGHTNODES". So the total number of nodes for the current level will be "("LEFTNODES" +"RIGHTNODES" )".

3. The maximum number of nodes among all levels will be the maximum width of the given tree.

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02 Approach

In Approach 1, to get the total number of nodes on each level, we used a recursive algorithm. That took cost O(N*H), where 'N' is the number of nodes in the given binary tree and 'H' is the height of the binary tree. But in this approach, our intuition is that as we know about preorder, postorder, or inorder traversal visits each element only once. We will be using any one of these three, and while calling the function, we will also pass the level corresponding to that node. And when we visit any node, we will increment the total number of nodes corresponding to the level of that node. The steps are as follows :

We will be using a HashMap to store the total number of nodes for each level, let's say "NODES-AT-LEVEL". The key of "NODES-AT-LEVEL" will represent the level, and the value corresponding to that key will represent the total number of nodes at that level.
We are going to apply preorder traversal for travelling each node, and our "LEVEL" will start from '0', steps are as follows:
- If the root is "NULL", we can not get any node for the current level.
- Else, we got a node for the current level, so increment the count.
  - "NODES-AT-LEVEL["LEVEL"]++”.
- Call the recursive function for left subtree and right subtree with "LEVEL+1".
Every value corresponding to the key of "NODES-AT-LEVEL" will represent the total number of nodes at that level in the given binary tree.
Return the maximum number of nodes in "NODES-AT-LEVEL". That will be our maximum width for the given binary tree.

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03 Approach

In this approach, we are going to use the level order traversal algorithm. While traversing in level order, we will be storing all the children of the nodes at the current level in the queue. And once we store all the children nodes for the current level, our queue will have all the next-level nodes. So, We can get the total number of nodes for the next level from the size of the queue. And we keep exploring and maintaining the maximum number of nodes in a level (i.e. the maximum size of the queue at any level) that we got till now. In the end, we have a maximum number of nodes that could be at any level. The steps are as follows:

We are going to use level order traversal, so push the root into the queue. And iterate through the queue until it becomes empty.
Get the queue's size at each level if the size of the queue is greater than the maximum number of nodes that we got till now, then we update our answer.
Push all the children of all the root nodes of the current level into the queue.
Return the final answer, which has stored the queue's maximum size at any level that will be the maximum width of the given binary tree.

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