Diameter of Binary Tree

Clarify the problem:

• The problem requires finding the diameter of a binary tree.
• The diameter of a binary tree is defined as the maximum number of edges between any two nodes in the tree.
• We need to implement a function that takes the root of a binary tree as input and returns the diameter of the tree.

Analyze the problem:

• Input: The root node of a binary tree.
• Output: The diameter of the binary tree (an integer).
• Constraints:
  • The binary tree can be empty.
  • The binary tree can have at most 10^4 nodes.

Design an algorithm:

• We can solve this problem using a recursive approach.
• We need to calculate the height of the left and right subtrees for each node.
• The diameter of the binary tree passing through a particular node is the sum of the heights of its left and right subtrees.
• We traverse the tree recursively and update the diameter at each node.

Explain your approach:

• We will implement a function called diameterOfBinaryTree that takes the root of a binary tree as input.
• If the root is None, we return 0 as the diameter.
• Otherwise, we recursively calculate the height of the left and right subtrees.
• We calculate the diameter passing through the current node by summing the heights of the left and right subtrees.
• We then recursively calculate the diameter for the left and right subtrees and take the maximum of these values.
• Finally, we return the maximum diameter.

Write clean and readable code:

python

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def diameterOfBinaryTree(root):
    def height(node):
        if not node:
            return 0
        left_height = height(node.left)
        right_height = height(node.right)
        return max(left_height, right_height) + 1

    def diameter(node):
        if not node:
            return 0
        left_diameter = diameter(node.left)
        right_diameter = diameter(node.right)
        return max(left_diameter, right_diameter, height(node.left) + height(node.right))

    return diameter(root)

Test your code:

python

# Test case 1
# Input:
#        1
#       / \
#      2   3
#     / \
#    4   5
# The diameter is the path 4 -> 2 -> 1 -> 3 with a length of 3.
# Expected output: 3
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
assert diameterOfBinaryTree(root) == 3

# Test case 2
# Input:
#        1
#       / \
#      2   3
#     / \     \
#    4   5     6
#   /         / \
#  7         8   9
# The diameter is the path 7 -> 4 -> 2 -> 1 -> 3 -> 6 -> 9 with a length of 6.
# Expected output: 6
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.left.left.left = TreeNode(7)
root.right.right = TreeNode(6)
root.right.right.left = TreeNode(8)
root.right.right.right = TreeNode(9)
assert diameterOfBinaryTree(root) == 6

Optimize if necessary:

• The solution already has an efficient time complexity, as we visit each node only once.
• There is no need for further optimization.

Handle error cases:

• The code handles the case where the root is None by returning 0 as the diameter.

Discuss complexity analysis:

• The time complexity of the solution is O(n), where n is the number of nodes in the binary tree. We visit each node once.
• The space complexity is O(h), where h is the height of the binary tree. This is the space required for the recursive call stack. In the worst case, the tree can be skewed, resulting in a space complexity of O(n), but in a balanced tree, the space complexity is O(log n).