DP on Trees Diameter of Tree in DSA C++ - Build from Scratch

DP on Trees: Diameter of Tree

📖 Scenario: You are working with a network of connected computers represented as a tree. Each connection is a direct link between two computers. You want to find the longest path between any two computers in this network. This longest path is called the diameter of the tree.

🎯 Goal: Build a C++ program that uses dynamic programming on trees to find the diameter of the given tree. You will represent the tree as an adjacency list, then use a depth-first search (DFS) to compute the diameter.

📋 What You'll Learn

Create an adjacency list to represent the tree with 5 nodes and given edges

Create a variable to store the diameter of the tree

Implement a DFS function that returns the longest path from a node and updates the diameter

Print the diameter of the tree after processing

💡 Why This Matters

🌍 Real World

Finding the diameter of a tree is useful in network design, biology (e.g., evolutionary trees), and communication systems to understand the longest distance between nodes.

💼 Career

Understanding tree diameter and dynamic programming on trees is important for software engineers working on graph algorithms, network optimization, and competitive programming.

1

Create the tree adjacency list

Create a vector of vectors of integers called tree with size 5 to represent the tree. Add edges between nodes as follows: 0-1, 0-2, 1-3, 1-4.

DSA C++

std::vector> tree(5);
// Your code here

Hint

Use std::vector> tree(5); to create the adjacency list. Add edges in both directions for an undirected tree.

2

Create a variable to store the diameter

Create an integer variable called diameter and initialize it to 0. This will store the longest path length found in the tree.

DSA C++

std::vector> tree(5);
tree[0].push_back(1);
tree[1].push_back(0);
tree[0].push_back(2);
tree[2].push_back(0);
tree[1].push_back(3);
tree[3].push_back(1);
tree[1].push_back(4);
tree[4].push_back(1);

int diameter = 0;
// Your code here

Hint

Use int diameter = 0; to create the variable.

3

Implement DFS to compute diameter

Write a recursive function called dfs that takes an integer node, an integer parent, and returns an integer. Inside, find the two longest depths from the current node's children. Update the global diameter with the sum of these two depths if it is larger. Return the longest depth plus one.

DSA C++

std::vector> tree(5);
tree[0].push_back(1);
tree[1].push_back(0);
tree[0].push_back(2);
tree[2].push_back(0);
tree[1].push_back(3);
tree[3].push_back(1);
tree[1].push_back(4);
tree[4].push_back(1);

int diameter = 0;

int dfs(int node, int parent) {
    // Your code here
}

Hint

Use two variables to track the longest and second longest depths from children. Update diameter with their sum. Return the longest depth plus one.

4

Run DFS and print the diameter

Call dfs starting from node 0 with parent -1. Then print the value of diameter.

DSA C++

std::vector> tree(5);
tree[0].push_back(1);
tree[1].push_back(0);
tree[0].push_back(2);
tree[2].push_back(0);
tree[1].push_back(3);
tree[3].push_back(1);
tree[1].push_back(4);
tree[4].push_back(1);

int diameter = 0;

int dfs(int node, int parent) {
    int max1 = 0, max2 = 0;
    for (int child : tree[node]) {
        if (child == parent) continue;
        int depth = dfs(child, node);
        if (depth > max1) {
            max2 = max1;
            max1 = depth;
        } else if (depth > max2) {
            max2 = depth;
        }
    }
    diameter = std::max(diameter, max1 + max2);
    return max1 + 1;
}

int main() {
    // Your code here
}

Hint

Call dfs(0, -1); to start DFS from node 0. Then print diameter using std::cout.

Practice

1. What does the diameter of a tree represent in graph theory?

easy

A. The longest path between any two nodes in the tree

B. The number of nodes in the tree

C. The height of the tree from the root

D. The number of leaf nodes in the tree

Solution

1. Step 1: Understand the definition of tree diameter

   The diameter is defined as the longest path between any two nodes in a tree, not necessarily passing through the root.

2. Step 2: Differentiate from other tree properties

   Height is the longest path from root to leaf, number of nodes and leaves are counts, not path lengths.

3. Final Answer:

   The longest path between any two nodes in the tree -> Option A

4. Quick Check:

   Diameter = longest path in tree [OK]

Hint: Diameter = longest path between any two nodes [OK]

Common Mistakes:

• Confusing diameter with height
• Thinking diameter counts nodes instead of path length
• Assuming diameter must pass through root

2. Which of the following is the correct way to declare a vector of vectors to represent an adjacency list for a tree in C++?

easy

A. std::vector<int> adj[n];

B. std::vector adj(n);

C. std::vector adj(n, std::vector<int>());

D. std::vector adj(n);

Solution

1. Step 1: Understand vector of vectors syntax

   Correct syntax uses std::vector adj(n); to create n empty vectors.

2. Step 2: Check other options for errors

   std::vector<int> adj[n]; uses array of vectors (valid but less modern), C misses template arguments, D misses template parameters.

3. Final Answer:

   std::vector adj(n); -> Option B

4. Quick Check:

   Correct vector of vectors syntax = std::vector adj(n); [OK]

Hint: Use std::vector adj(n); for adjacency list [OK]

Common Mistakes:

• Missing template parameters in vector
• Using C-style arrays instead of vectors
• Incorrect initialization syntax

3. Consider the following C++ code snippet to find the diameter of a tree using DFS and DP. What will be the output if the tree edges are: (0-1), (1-2), (1-3)?

std::vector adj(4);
adj[0].push_back(1); adj[1].push_back(0);
adj[1].push_back(2); adj[2].push_back(1);
adj[1].push_back(3); adj[3].push_back(1);

int diameter = 0;

int dfs(int node, int parent, const std::vector& adj, int& diameter) {
    int max1 = 0, max2 = 0;
    for (int child : adj[node]) {
        if (child == parent) continue;
        int height = dfs(child, node, adj, diameter) + 1;
        if (height > max1) {
            max2 = max1;
            max1 = height;
        } else if (height > max2) {
            max2 = height;
        }
    }
    diameter = std::max(diameter, max1 + max2);
    return max1;
}

dfs(0, -1, adj, diameter);
std::cout << diameter << std::endl;

medium

A. 3

B. 1

C. 0

D. 2

Solution

1. Step 1: Understand the tree structure and DFS traversal

   The tree edges form a structure: 0-1 connected, 1 connected to 2 and 3. The longest path is from node 2 to node 3 through 1, length 2 edges.

2. Step 2: Calculate diameter from DFS logic

   DFS calculates max1 and max2 heights from children. For node 1, max1=1 (to 2), max2=1 (to 3), so diameter = max(0,1+1)=2.

3. Final Answer:

   2 -> Option D

4. Quick Check:

   Longest path length = 2 edges [OK]

Hint: Diameter = sum of two longest child heights at any node [OK]

Common Mistakes:

• Counting nodes instead of edges for diameter
• Not updating diameter correctly
• Confusing parent and child in DFS

4. The following code attempts to compute the diameter of a tree but produces incorrect results. What is the main bug?

int dfs(int node, int parent, const std::vector& adj, int& diameter) {
    int maxHeight = 0;
    for (int child : adj[node]) {
        if (child == parent) continue;
        int height = dfs(child, node, adj, diameter) + 1;
        if (height > maxHeight) {
            maxHeight = height;
            diameter = std::max(diameter, maxHeight);
        }
    }
    return maxHeight;
}

medium

A. DFS does not return any value

B. Parent node is not skipped in DFS

C. Diameter is updated only with maxHeight, missing sum of two heights

D. Height is not incremented correctly

Solution

1. Step 1: Analyze diameter update logic

   The code updates diameter only with maxHeight, which is the longest single path from current node, but diameter requires sum of two longest paths through the node.

2. Step 2: Identify missing second longest path

   Without tracking two longest heights, the code misses paths passing through the node combining two children, leading to incorrect diameter.

3. Final Answer:

   Diameter is updated only with maxHeight, missing sum of two heights -> Option C

4. Quick Check:

   Diameter needs sum of two longest child heights [OK]

Hint: Track two longest heights to update diameter correctly [OK]

Common Mistakes:

• Updating diameter with only one child's height
• Not skipping parent node in DFS
• Forgetting to increment height by 1

5. Given a tree with 6 nodes and edges: (0-1), (1-2), (1-3), (3-4), (4-5), what is the diameter of the tree? Use the DP on Trees approach where diameter is the longest path between any two nodes.

hard

A. 5

B. 4

C. 3

D. 6

Solution

1. Step 1: Visualize the tree and identify longest path

   The tree edges form paths: 0-1-3-4-5 and 2 connected to 1. The longest path is from node 2 to node 5 passing through 1,3,4 with length 5 edges.

2. Step 2: Calculate diameter using DP on Trees logic

   At node 1, two longest child heights are 1 (to 2) and 3 (to 5). Diameter = max(diameter, 1+3) = 4, but the longest path is from 2 to 5 which is 4 edges plus 1 edge from 0 to 1 is not part of longest path. Actually, the longest path is from node 2 to node 5 passing through 1,3,4 with length 5 edges.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Longest path length = 5 edges [OK]

Hint: Diameter = sum of two longest child heights at a node [OK]

Common Mistakes:

• Counting nodes instead of edges for diameter
• Ignoring one branch in longest path
• Miscounting edges in path