Bird
Raised Fist0
DSA Pythonprogramming~3 mins

Why DP on Trees Diameter of Tree in DSA Python?

Choose your learning style10 modes available
The Big Idea

Discover how to find the longest path in a tree without checking every route by hand!

The Scenario

Imagine you have a big family tree drawn on paper. You want to find the longest path between any two family members, but you have to check every possible path by hand.

The Problem

Checking every path manually is slow and confusing. You might miss some paths or repeat checking the same parts again and again. It's easy to get lost and make mistakes.

The Solution

Dynamic Programming on Trees helps by breaking the problem into smaller parts. It remembers the longest paths from each node so it doesn't have to repeat work. This way, it quickly finds the longest path in the whole tree.

Before vs After
✗ Before
def longest_path(tree):
    # Check all paths manually
    max_length = 0
    for start in tree:
        for end in tree:
            length = find_path_length(start, end)
            max_length = max(max_length, length)
    return max_length
✓ After
def diameter_of_tree(node):
    def dfs(current):
        max1 = max2 = 0
        for child in current.children:
            length = dfs(child) + 1
            if length > max1:
                max2 = max1
                max1 = length
            elif length > max2:
                max2 = length
        diameter[0] = max(diameter[0], max1 + max2)
        return max1
    diameter = [0]
    dfs(node)
    return diameter[0]
What It Enables

This method lets you find the longest path in a tree quickly and reliably, even for very large trees.

Real Life Example

In a network of computers connected like a tree, finding the longest path helps understand the worst-case delay between two computers.

Key Takeaways

Manual checking of all paths is slow and error-prone.

DP on Trees breaks the problem into smaller parts and remembers results.

This approach finds the longest path (diameter) efficiently.

Practice

(1/5)
1. What is the diameter of a tree in terms of its nodes?
easy
A. The total number of nodes in the tree
B. The number of leaf nodes in the tree
C. The longest path between any two nodes in the tree
D. The height of the tree from the root node

Solution

  1. Step 1: Understand the definition of tree diameter

    The diameter is the longest path between any two nodes, not just from the root or leaf count.
  2. Step 2: Compare options with definition

    Only The longest path between any two nodes in the tree correctly describes the longest path between any two nodes.
  3. Final Answer:

    The longest path between any two nodes in the tree -> Option C
  4. Quick Check:

    Diameter = longest path [OK]
Hint: Diameter = longest path between any two nodes [OK]
Common Mistakes:
  • Confusing diameter with tree height
  • Thinking diameter is number of leaves
  • Assuming diameter is total nodes count
2. Which of the following is the correct Python syntax to declare a recursive function to find the diameter of a tree?
easy
A. def diameter(node):
B. function diameter(node) {
C. def diameter[node]:
D. diameter = lambda node:

Solution

  1. Step 1: Identify correct Python function syntax

    Python functions use 'def' keyword followed by function name and parentheses with parameters.
  2. Step 2: Check each option

    def diameter(node): uses correct Python syntax. function diameter(node) { is JavaScript style. def diameter[node]: uses wrong brackets. diameter = lambda node: is incomplete lambda syntax.
  3. Final Answer:

    def diameter(node): -> Option A
  4. Quick Check:

    Python function syntax = def name(params): [OK]
Hint: Python functions start with def and parentheses [OK]
Common Mistakes:
  • Using curly braces instead of colon
  • Using square brackets for parameters
  • Incorrect lambda syntax for recursion
3. Given the following tree edges and code snippet, what is the diameter of the tree?
edges = [(1,2), (2,3), (2,4), (4,5)]

# Code snippet
from collections import defaultdict

tree = defaultdict(list)
for u,v in edges:
    tree[u].append(v)
    tree[v].append(u)

max_diameter = 0
def dfs(node, parent):
    global max_diameter
    max1 = max2 = 0
    for child in tree[node]:
        if child == parent:
            continue
        depth = dfs(child, node)
        if depth > max1:
            max2 = max1
            max1 = depth
        elif depth > max2:
            max2 = depth
    max_diameter = max(max_diameter, max1 + max2)
    return max1 + 1

dfs(1, -1)
print(max_diameter)
medium
A. 3
B. 4
C. 5
D. 2

Solution

  1. Step 1: Build the tree and understand structure

    The tree edges form: 1-2-3, 2-4-5. The longest path is from node 3 to node 5 via 2 and 4.
  2. Step 2: Calculate longest path length

    Path 3->2->4->5 has 3 edges, so diameter is 3 edges = 4 nodes in path length. The code counts edges, so max_diameter = 3 edges.
  3. Final Answer:

    3 -> Option A
  4. Quick Check:

    Longest path edges = 3 [OK]
Hint: Diameter = max sum of two largest depths at any node [OK]
Common Mistakes:
  • Counting nodes instead of edges for diameter
  • Ignoring parent check in DFS
  • Mixing up max1 and max2 updates
4. The following code attempts to find the diameter of a tree but produces incorrect output. What is the bug?
def dfs(node, parent):
    max_depth = 0
    for child in tree[node]:
        if child != parent:
            depth = dfs(child, node)
            if depth > max_depth:
                max_depth = depth
    return max_depth + 1

max_diameter = 0
def find_diameter(node, parent):
    global max_diameter
    depths = []
    for child in tree[node]:
        if child != parent:
            depths.append(dfs(child, node))
    if len(depths) >= 2:
        depths.sort(reverse=True)
        max_diameter = max(max_diameter, depths[0] + depths[1])
    for child in tree[node]:
        if child != parent:
            find_diameter(child, node)

find_diameter(1, -1)
print(max_diameter)
medium
A. The recursive call to find_diameter is missing
B. max_diameter is not updated globally
C. Sorting depths in descending order is incorrect
D. Only updates max_diameter when len(depths) >= 2, missing updates for nodes with fewer than 2 children

Solution

  1. Step 1: Identify the conditional update bug

    The code only updates max_diameter if len(depths) >= 2, but to mimic standard algorithm, it should always update using the two largest depths (with second as 0 if fewer children).
  2. Step 2: Understand impact on chain trees

    In chain trees, nodes often have at most 1 child (excluding parent), so the condition skips updates, leaving max_diameter = 0 incorrectly.
  3. Final Answer:

    Only updates max_diameter when len(depths) >= 2, missing updates for nodes with fewer than 2 children -> Option D
  4. Quick Check:

    Pad missing depths with 0 for diameter update [OK]
Hint: Always update diameter with top two downward depths (pad 0 if needed) [OK]
Common Mistakes:
  • Not updating global max_diameter
  • Incorrect sorting order of depths
  • Missing recursive calls in diameter function
5. You want to find the diameter of a weighted tree where edges have different lengths. Which modification to the standard DP on trees diameter algorithm is correct? A weighted tree is represented as adjacency list with tuples (neighbor, weight).
tree = {
  1: [(2, 3), (3, 2)],
  2: [(1, 3), (4, 4)],
  3: [(1, 2)],
  4: [(2, 4)]
}

# Which code snippet correctly computes diameter?
hard
A. Use BFS twice ignoring weights to find diameter
B. Use dfs to return max weighted depth, update diameter by sum of top two weighted depths
C. Ignore weights and count edges only for diameter
D. Sum all weights in the tree and divide by 2

Solution

  1. Step 1: Understand weighted tree diameter

    Diameter is longest weighted path, so weights must be included in depth calculations.
  2. Step 2: Modify DFS to return max weighted depth and update diameter

    DFS returns max weighted depth from node; diameter updated by sum of two largest weighted depths.
  3. Final Answer:

    Use dfs to return max weighted depth, update diameter by sum of top two weighted depths -> Option B
  4. Quick Check:

    Weighted diameter = sum of max weighted depths [OK]
Hint: Include edge weights in depth sums for weighted diameter [OK]
Common Mistakes:
  • Ignoring weights and treating as unweighted
  • Using BFS without weights for weighted diameter
  • Summing all weights instead of longest path
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard