DP on Trees Diameter of Tree in DSA C - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the diameter of a tree?

The diameter of a tree is the length of the longest path between any two nodes in the tree.

intermediate

How can dynamic programming help in finding the diameter of a tree?

Dynamic programming helps by storing the longest path from each node to its descendants, avoiding repeated calculations and efficiently computing the diameter.

beginner

In the DP approach for tree diameter, what does the 'height' of a node represent?

The height of a node is the length of the longest path from that node down to a leaf node in its subtree.

intermediate

Why do we consider the two longest heights among children when computing the diameter at a node?

Because the longest path through that node can be formed by joining the two longest paths from its children, passing through the node itself.

beginner

What is the time complexity of finding the diameter of a tree using DP?

The time complexity is O(N), where N is the number of nodes, because each node and edge is visited once during DFS traversal.

What does the diameter of a tree represent?

A Longest path between any two nodes

B Number of nodes in the tree

C Height of the tree

D Number of leaves in the tree

The diameter is the longest path between any two nodes in the tree.

In DP on trees, what is the 'height' of a node?

A Number of edges in the tree

B Number of children of the node

C Distance from root to the node

D Longest path from node to a leaf in its subtree

Height is the longest path from the node down to a leaf in its subtree.

Why do we use two longest heights from children to compute diameter at a node?

A To count the number of children

B To find the shortest path

C To find the longest path passing through the node

D To calculate the node's degree

The longest path through the node is formed by joining the two longest child paths.

What traversal method is commonly used to compute diameter using DP on trees?

A Breadth-first search

B Depth-first search

C Level order traversal

D Inorder traversal

Depth-first search is used to compute heights and diameters efficiently.

What is the time complexity of the DP approach to find tree diameter?

A O(N)

B O(N^2)

C O(N log N)

D O(log N)

Each node is visited once, so the complexity is O(N).

Explain how dynamic programming is applied to find the diameter of a tree.

Think about how to combine child heights to find longest path through a node.

Describe the steps to compute the diameter of a tree using a single DFS traversal.

Focus on how to gather information bottom-up during DFS.

Practice

1. What is the diameter of a tree in the context of dynamic programming on trees?

easy

A. The height of the tree from the root

B. The number of nodes in the tree

C. The sum of all edge weights in the tree

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

Solution

1. Step 1: Understand the definition of tree diameter

   The diameter is the longest path between any two nodes in the tree, not just from the root.

2. Step 2: Differentiate from other tree properties

   Height is from root to leaf, sum of edges is total weight, and node count is size, none represent longest path.

3. Final Answer:

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

4. Quick Check:

   Diameter = Longest path [OK]

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

Common Mistakes:

• Confusing diameter with height
• Thinking diameter is number of nodes
• Assuming diameter is sum of all edges

2. Which of the following is the correct way to declare an adjacency list for a tree with n nodes in C?

easy

A. int adj[n][n];

B. int* adj[n];

C. int adj[n];

D. int adjList[n][2];

Solution

1. Step 1: Understand adjacency list structure

   An adjacency list uses an array of pointers to lists (or arrays) of neighbors, so int* adj[n]; is correct.

2. Step 2: Differentiate from other declarations

   int adj[n][n]; is adjacency matrix, int adj[n]; is a simple array, int adjList[n][2]; is edge list, not adjacency list.

3. Final Answer:

   int* adj[n]; -> Option B

4. Quick Check:

   Adjacency list uses array of pointers [OK]

Hint: Adjacency list uses array of pointers to neighbors [OK]

Common Mistakes:

• Using 2D array as adjacency list
• Using simple array for adjacency
• Confusing edge list with adjacency list

3. Consider the following C code snippet to find the diameter of a tree. What will be the output if the tree has edges: 1-2, 2-3, 2-4?

int diameter = 0;
int dfs(int node, int parent, vector<int> adj[]) {
    int max1 = 0, max2 = 0;
    for (int child : adj[node]) {
        if (child == parent) continue;
        int height = dfs(child, node, adj) + 1;
        if (height > max1) {
            max2 = max1;
            max1 = height;
        } else if (height > max2) {
            max2 = height;
        }
    }
    diameter = max(diameter, max1 + max2);
    return max1;
}
// Call dfs(1, -1, adj) and then print diameter;

medium

A. 2

B. 3

C. 1

D. 4

Solution

1. Step 1: Trace DFS from node 1

   From node 1, child is 2. DFS(2) explores children 3 and 4. Heights from 3 and 4 are 1 each.

2. Step 2: Calculate diameter updates

   At node 2, max1=1, max2=1, diameter updated to 2. At node 1, max1=2, max2=0, diameter remains 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Longest path length = 2 edges [OK]

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

Common Mistakes:

• Counting nodes instead of edges
• Not updating diameter correctly
• Ignoring parent to avoid cycles

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

int diameter = 0;
int dfs(int node, int parent, vector<int> adj[]) {
    int max1 = 0, max2 = 0;
    for (int child : adj[node]) {
        int height = dfs(child, node, adj) + 1;
        if (height > max1) {
            max2 = max1;
            max1 = height;
        } else if (height > max2) {
            max2 = height;
        }
    }
    diameter = max(diameter, max1 + max2);
    return max1;
}
// Called with dfs(1, -1, adj);

medium

A. Incorrect initialization of diameter variable

B. Returning max2 instead of max1 from DFS

C. Missing check to skip parent node in DFS

D. Not adding 1 to height in DFS call

Solution

1. Step 1: Identify missing parent check in DFS loop

   The code does not skip the parent node, causing infinite recursion or wrong heights.

2. Step 2: Explain impact of missing check

   Without skipping parent, DFS revisits parent, causing incorrect diameter calculation or stack overflow.

3. Final Answer:

   Missing check to skip parent node in DFS -> Option C

4. Quick Check:

   Skip parent to avoid cycles [OK]

Hint: Always skip parent node in DFS to prevent cycles [OK]

Common Mistakes:

• Forgetting to skip parent node
• Returning wrong height value
• Incorrect diameter initialization

5. Given a weighted tree with edges: (1-2, 3), (2-3, 4), (2-4, 2), how would you modify the diameter calculation to find the longest weighted path? Choose the correct approach.

hard

A. Add edge weights to height calculations and update diameter with weighted sums

B. Ignore weights and use number of edges as diameter

C. Multiply edge weights instead of adding

D. Use BFS twice without considering weights

Solution

1. Step 1: Incorporate edge weights in DFS height calculation

   Instead of adding 1 for each edge, add the actual edge weight to the height returned by DFS.

2. Step 2: Update diameter with sum of two largest weighted heights

   Keep track of two largest weighted heights from children and update diameter with their sum.

3. Final Answer:

   Add edge weights to height calculations and update diameter with weighted sums -> Option A

4. Quick Check:

   Weighted diameter sums edge weights [OK]

Hint: Add weights to heights, sum two largest for diameter [OK]

Common Mistakes:

• Ignoring weights in diameter calculation
• Multiplying weights instead of adding
• Using unweighted BFS for weighted tree