DP on Trees Diameter of Tree in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the adjacency list for the tree.

DSA C++

vector<int> adj[[1]];

Drag options to blanks, or click blank then click option'

An

Bdiameter

Cedges

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using number of edges instead of nodes for adjacency list size.

The adjacency list size should be equal to the number of nodes 'n' in the tree.

2fill in blank

medium

Complete the code to update the diameter during DFS traversal.

DSA C++

diameter = max(diameter, dist1 + dist2 + [1]);

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A0

B1

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Adding 1 or 2 instead of 0, causing off-by-one errors.

No additional value is needed because dist1 and dist2 represent the lengths of the two longest downward paths from the current node, and their sum gives the diameter through this node.

3fill in blank

hard

Fix the error in the DFS function to correctly compute the longest path from the current node.

DSA C++

int dfs(int node, int parent) {
    int max_depth = 0;
    for (int child : adj[node]) {
        if (child == [1]) continue;
        int depth = dfs(child, node) + 1;
        max_depth = max(max_depth, depth);
    }
    return max_depth;
}

Drag options to blanks, or click blank then click option'

A-1

Bnode

Cchild

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Skipping the current node or child instead of the parent.

We skip the parent node to avoid revisiting it and causing infinite recursion.

4fill in blank

hard

Fill both blanks to update the two longest depths from children in the DFS.

DSA C++

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

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Amax1

Bmax2

Cdepth

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping max1 and max2 incorrectly or using wrong variables.

We compare depth with max1, then assign max2 to old max1 before updating max1.

5fill in blank

hard

Fill all three blanks to implement the diameter calculation using DFS.

DSA C++

int diameter = 0;

int dfs(int node, int parent) {
    int max1 = 0, max2 = 0;
    for (int child : adj[node]) {
        if (child == parent) continue;
        int depth = dfs(child, node) + [1];
        if (depth > max1) {
            max2 = max1;
            max1 = depth;
        } else if (depth > max2) {
            max2 = depth;
        }
    }
    diameter = max(diameter, max1 + max2 + [2]);
    return max1 + [3];
}

Drag options to blanks, or click blank then click option'

A1

B0

C-1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Adding 1 or 2 incorrectly causing off-by-one errors.

Add 1 for edge length when going down, add 0 when updating diameter (already counted), and add 0 when returning max1.

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