DP on Trees Diameter of Tree in DSA C++ - Time & Space Complexity

Time Complexity: DP on Trees Diameter of Tree

O(n)

Understanding Time Complexity

We want to understand how long it takes to find the longest path in a tree using dynamic programming.

How does the time grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

int treeDiameter(vector<vector<int>>& adj) {
    int diameter = 0;
    dfs(0, -1, adj, diameter);
    return diameter;
}
    

This code finds the diameter (longest path) of a tree using a depth-first search and dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The depth-first search (dfs) visits each node once and explores its children.
• How many times: Each node is visited exactly one time, and each edge is checked twice (once from each node).

How Execution Grows With Input

As the number of nodes grows, the number of operations grows roughly in proportion to the number of nodes.

Input Size (n)	Approx. Operations
10	About 18 (edges checked twice)
100	About 198
1000	About 1998

Pattern observation: The operations grow linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the diameter grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The diameter calculation takes quadratic time because it checks all pairs of nodes."

[OK] Correct: The code uses a single depth-first search that visits each node once, so it does not check all pairs explicitly.

Interview Connect

Understanding this linear time approach shows you can efficiently solve tree problems using recursion and dynamic programming.

Self-Check

"What if the tree was represented as an adjacency matrix instead of adjacency lists? How would the time complexity change?"

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