Program to find the maximum width of a binary tree in Python

Suppose we have a binary tree, we have to find the maximum width of any level in the tree. Here the width of a level is the number of nodes that can hold between the leftmost node and the rightmost node.

So, if the input is like 

then the output will be 2

To solve this, we will follow these steps−

• create a map d, to hold minimum and maximum values, minimum is initially infinity and maximum is 0

• Define a function dfs() . This will take root, pos := 0, depth := 0

• if root is null, then o return

• d[depth, 0] = minimum of d[depth,0] and pos

• d[depth, 1] = maximum of d[depth,1] and pos

• dfs(left of node, 2*pos, depth+1)

• dfs(right of node, 2*pos+1, depth+1)

• From the main method, do the following−

• dfs(root)

• mx:= 0

• for each min-max pairs in list of all values of d, do

  • left := min, right := max

  • mx:= maximum of mx, right-lelf+1

• return mx

Let us see the following implementation to get better understanding −

Example

 Live Demo

from collections import defaultdict
   class TreeNode:
      def __init__(self, data, left = None, right = None):
         self.data = data
         self.left = left
         self.right = right
class Solution:
   def solve(self, root):
   d=defaultdict(lambda: [1e9,0])
   def dfs(node, pos=0, depth=0):
      if not node:
         return
      d[depth][0]=min(d[depth][0],pos)
      d[depth][1]=max(d[depth][1],pos)
      dfs(node.left,2*pos,depth+1)
      dfs(node.right,2*pos+1,depth+1)
   dfs(root)
   mx=0
   for interval in d.values():
      l,r=interval
      mx=max(mx,r-l+1)
   return mx

ob = Solution()
root = TreeNode(5)
root.left = TreeNode(1)
root.right = TreeNode(9)
root.right.left = TreeNode(7)
root.right.right = TreeNode(10)
root.right.left.left = TreeNode(6)
root.right.left.right = TreeNode(8)
print(ob.solve(root))

Input

root = TreeNode(5)
root.left = TreeNode(1)
root.right = TreeNode(9)
root.right.left = TreeNode(7)
root.right.right = TreeNode(10)
root.right.left.left = TreeNode(6)
root.right.left.right = TreeNode(8)

Output

2