Grind 75 - 21 - Diameter of Binary Tree

LeetCode Problem: Diameter of a Binary Tree

• LeetCode Number: 543
• Difficulty Level: Easy

Explanation of the Question:

• Problem Statement: Given the root of a binary tree, return the length of the diameter of the tree.
• Definition: The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
• Input: Root of a binary tree.
• Output: An integer, representing the diameter of the tree.
• Sample Data:Input: [1,2,3,4,5]
• Output: 3

Questions to Be Asked to the Interviewer:

• Can the tree be empty?
• Is the tree guaranteed to be balanced?
• Can I assume the nodes' values are irrelevant to the solution?

Edge Cases:

• If the tree is empty, the diameter is 0.
• If the tree only consists of a single node, the diameter is also 0.

Explain the Available Approaches:

• Approach 1: Traverse the tree, calculating the depth of left and right children recursively, and use the sum of both sides plus 1 as the diameter.
• Approach 2: Similar to approach 1 but with memoization to avoid recalculating depths.

My Approach:

• Design: Use a recursive depth-first search (DFS) to calculate the depth of left and right children. At each node, the maximum diameter is updated using the sum of left and right depths.
• Classification: This falls under Tree Traversal algorithms (specifically, Depth-First Search) and the use of Binary Trees as a data structure.

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


class Solution:
? ? def diameterOfBinaryTree(self, root: TreeNode) -> int:
? ? ? ? def depth(node):
? ? ? ? ? ? nonlocal diameter
? ? ? ? ? ? if not node:
? ? ? ? ? ? ? ? return 0
? ? ? ? ? ? left_depth = depth(node.left)
? ? ? ? ? ? right_depth = depth(node.right)
? ? ? ? ? ? diameter = max(diameter, left_depth + right_depth)
? ? ? ? ? ? return max(left_depth, right_depth) + 1
? ? ? ??
? ? ? ? diameter = 0
? ? ? ? depth(root)
? ? ? ? return diameter
        

Explain the Solution to a Non-Programmer :

• Imagine a Tree: Think of the binary tree as a family tree with a root, representing the top ancestor, and branches, representing children and grandchildren.
• Longest Path: We want to find the longest path between any two people in this family tree. This path could be between siblings, cousins, etc.
• Visit Everyone: We go through the tree, visiting everyone's left and right "children" (or branches).
• Measure the Branches: At each person, we measure how long the branches are to the left and right.
• Longest Connection: The longest connection between any two people will be the sum of these left and right branches somewhere in the tree.
• Result: We return this longest connection as the "diameter" of the family tree.

Code in Detail:

• Define Tree Structure: A class TreeNode is defined to represent the nodes of the binary tree.
• Diameter Function: The main function diameterOfBinaryTree takes the root of the tree as input.
• Depth Function: A nested function depth is defined to recursively find the depth of a node in the tree.If the node is None, return 0 (base case).
• Recursively find the depths of the left and right children.
• Update the diameter using the sum of left and right depths.
• Return the maximum of left and right depths plus 1.
• Initialize Diameter: Outside of the depth function, initialize diameter as 0.
• Calculate Diameter: Call the depth function starting from the root to calculate the diameter.
• Return Result: Return the final value of diameter.

Pattern of this Problem:

• This problem represents a common pattern of calculating a property in a binary tree using depth-first search.
• It involves recursive traversal and calculation of a value (in this case, diameter) based on properties of left and right children.

Big O Notation:

• Time Complexity: The time complexity of this code is O(n) -O(n), where n
• n is the number of nodes in the binary tree. This is because the code visits every node in the tree exactly once.
• Space Complexity: The space complexity is O(h)- >O(h), where ?
• h is the height of the binary tree. This accounts for the recursive call stack, which can be as deep as the height of the tree.

Points to Remember to Solve this Problem in the Future:

• Consider a recursive approach to traverse the binary tree.
• Think about what values need to be calculated at each node (in this case, the depth) and how they contribute to the final result (diameter).
• Be careful with the base case and the way recursion is handled.

Code Line by Line with Some Sample Data:

Here's a step-by-step walkthrough of the code:

1. Initialize Diameter: diameter = 0
2. Define Depth Function: A recursive function to calculate the depth of the tree.
3. If Node is None: Return 0 (base case).
4. Calculate Left Depth: If the left child exists, recursively call depth(node.left).
5. Calculate Right Depth: If the right child exists, recursively call depth(node.right).
6. Update Diameter: diameter = max(diameter, left_depth + right_depth).
7. Return Depth of Current Node: Return max(left_depth, right_depth) + 1.
8. Start the Process: Call depth(root).
9. Return the Result: Return the value of diameter.

Sample Data: Let's say we have a binary tree as follows:

? ? 1
? ?/ \
? 2? ?3
?/ \
4? ?5


        

Key Syntax Used in this Solution :

• Defining a class for Tree Node: class TreeNode:
• Recursive function calls: Using recursion to traverse the tree.
• Updating a non-local variable within a nested function: Using the nonlocal keyword (Python 3) .