Program to find top view of a binary tree in Python

Suppose we have a binary tree, we have to find the top view of the tree, they will be sorted left−to−right.

So, if the input is like image, then the output will be [3, 5, 8, 6, 9], as 3 is above 2 and 5 is above 7 so they are not visible.

To solve this, we will follow these steps −

• view := a new empty map

• q := a double ended queue

• insert pair (root, 0) at the end of q

• start := inf, end := −inf

• while q is not empty, do

  • (node, coord) := left element of q, then remove left element of q

  • start := minimum of start and coord

  • end := maximum of end and coord

  • if coord is not in view, then

    • view[coord] := value of node

  • if left of node is not null, then

    • insert (left of node, coord − 1) at the end of q

  • if right of node is not null, then

    • insert (right of node, coord + 1) at the end of q

• res := a new list

• for i in range start to end, do

  • if i is in view, then

    • insert view[i] at the end of res

• return res

Let us see the following implementation to get better understanding −

Example

 Live Demo

from collections import deque
class TreeNode:
   def __init__(self, data, left = None, right = None):
      self.val = data
      self.left = left
      self.right = right
class Solution:
   def solve(self, root):
      view = {}
      q = deque()
      q.append((root, 0))
      start = float("inf")
      end = float("-inf")
      while q:
         node, coord = q.popleft()
         start = min(start, coord)
         end = max(end, coord)
            if coord not in view:
               view[coord] = node.val
            if node.left:
               q.append((node.left, coord - 1))
            if node.right:
               q.append((node.right, coord + 1))
         res = []
         for i in range(start, end + 1):
            if i in view:
               res.append(view[i])
         return res
ob = Solution()
root = TreeNode(5)
root.left = TreeNode(3)
root.right = TreeNode(8)
root.right.left = TreeNode(7)
root.right.right = TreeNode(6)
root.right.left.left = TreeNode(2)
root.right.right.right = TreeNode(9)
print(ob.solve(root))

Input

root = TreeNode(5) root.left = TreeNode(3) root.right = TreeNode(8)
root.right.left = TreeNode(7) root.right.right = TreeNode(6)
root.right.left.left = TreeNode(2) root.right.right.right =
TreeNode(9)

Output

[3, 5, 8, 6, 9]