Diameter of Binary Tree in Python

Suppose we have a binary tree; we have to compute the length of the diameter of the tree. The diameter of a binary tree is actually the length of the longest path between any two nodes in a tree. This path not necessarily pass through the root. So if the tree is like below, then the diameter will be 3.as the length of the path [4,2,1,3] or [5,2,1,3] is 3

To solve this, we will follow these steps −

• We will use the dfs to find the diameter, set answer := 0
• call the dfs function with the root dfs(root)
• dfs will work like below dfs(node)
• if node is not present, then return 0
• left := dfs(left subtree of root), and right := dfs(right subtree of root)
• answer := max of answer and left + right
• return max of left + 1 and right + 1

Example

Let us see the following implementation to get better understanding −

class TreeNode:
   def __init__(self, data, left = None, right = None):
      self.data = data
      self.left = left
      self.right = right
def insert(temp,data):
   que = []
   que.append(temp)
   while (len(que)):
      temp = que[0]
      que.pop(0)
      if (not temp.left):
         temp.left = TreeNode(data)
         break
      else:
         que.append(temp.left)
      if (not temp.right):
         temp.right = TreeNode(data)
         break
      else:
         que.append(temp.right)
def make_tree(elements):
   Tree = TreeNode(elements[0])
   for element in elements[1:]:
      insert(Tree, element)
   return Tree
class Solution(object):
   def diameterOfBinaryTree(self, root):
      """
      :type root: TreeNode
      :rtype: int
      """
      self.ans = 0
      self.dfs(root)
      return self.ans
   def dfs(self, node):
      if not node:
         return 0
      left = self.dfs(node.left)
      right = self.dfs(node.right)
      self.ans =max(self.ans,right+left)
      return max(left+1,right+1)
root = make_tree([1,2,3,4,5])
ob1 = Solution()
print(ob1.diameterOfBinaryTree(root))

Input

[1,2,3,4,5]

Output

3