Skip to main content
Code ScholarsCodeScholars

Trees and Balanced Search Structures · Trees

Tree Data Structure

Trees model hierarchy with roots, children, parents, leaves, subtrees, height, and depth.

Enroll NowText UsContact Code ScholarsFull DSA Course

Student Focus

We anchor tree ideas in visual structures before writing recursive code.

Guided Lesson Notes

Understanding Tree Data Structure

Tree Data Structure focuses on hierarchical relationships, recursive subproblems, and shape rules that control performance. Trees model hierarchy with roots, children, parents, leaves, subtrees, height, and depth.

Model: draw nodes as parent-child relationships; every node is the root of a smaller tree with the same rules.

Invariant: the shape or ordering rule must hold at every node, not just near the root.

Implementation: recursive traversal, iterative queues for levels, search paths, rotations, splits, merges, or range-query recursion.

Use Tree Data Structure when the problem describes hierarchy, ancestry, subtrees, intervals, prefixes, sorted dynamic data, or range queries.

Visual Model

A small picture for Tree Data Structure

rootLRABCD

Key Ideas

  • Tree vocabulary
  • Recursive structure
  • Hierarchical modeling

Practice Prompts

  • Label root, leaves, height, and depth in a sample tree.
  • Represent a folder system or tournament bracket as a tree.

Vocabulary

Terms students should be able to say clearly

Root

The top node where tree reasoning begins.

Leaf

A node with no children.

Height

The length of the longest downward path from a node.

Subtree

A node together with all descendants below it.

Traversal

A systematic order for visiting nodes.

Balance

A shape condition that keeps paths from becoming too long.

Worked Example

Worked example: tracing Tree Data Structure

Trace a small input, focusing on tree vocabulary.

  1. 1Write a small input where Tree Data Structure is clearly useful.
  2. 2Label the part of the input related to Tree vocabulary.
  3. 3Perform one operation and explain which invariant is still true afterward.
  4. 4Run a second operation that touches an edge case, such as an empty side, duplicate value, boundary index, disconnected vertex, or tie.
  5. 5Finish by saying which operation dominates the runtime and why.

Complexity Check

Costs students should be able to explain

OperationTypical CostReason
TraversalO(n)Visiting every node takes time proportional to the number of nodes.
Balanced searchO(log n)Balanced height keeps search paths short.
Unbalanced searchO(n)A poor shape can behave like a linked list.

Common Mistakes

What to watch while practicing

  • Coding before drawing the structure or state changes.
  • Forgetting the invariant that makes the algorithm correct.
  • Testing only the sample input and skipping boundary cases.
  • Giving Big-O without explaining which operation dominates the work.

Interview-Style Coding Problem

Tree Data Structure interview problem: Diameter of a Tree

Interview medium

Problem

Given an undirected tree, find the number of edges on the longest path between any two nodes.

Input

The first line contains n. The next n - 1 lines contain edges u v using 1-based node labels.

Output

Print the tree diameter in edges.

Sample Input

6
1 2
2 3
2 4
4 5
5 6

Sample Output

4

Why the sample works

One longest path is 3-2-4-5-6, which uses four edges.

Approach

  1. 1Run BFS from any node to find a farthest endpoint.
  2. 2Run BFS again from that endpoint.
  3. 3The largest distance in the second BFS is the diameter.
  4. 4This works because a farthest node from any start is an endpoint of a longest path.

O(n) time and O(n) extra space.

Java Solution

import java.util.*;

public class Main {
  static int[] bfs(List<Integer>[] graph, int start) {
    int n = graph.length - 1;
    int[] dist = new int[n + 1];
    Arrays.fill(dist, -1);

    Queue<Integer> q = new ArrayDeque<>();
    q.add(start);
    dist[start] = 0;

    while (!q.isEmpty()) {
      int node = q.remove();
      for (int next : graph[node]) {
        if (dist[next] == -1) {
          dist[next] = dist[node] + 1;
          q.add(next);
        }
      }
    }

    int farthest = start;
    for (int i = 1; i <= n; i++) {
      if (dist[i] > dist[farthest]) farthest = i;
    }

    return new int[] {farthest, dist[farthest]};
  }

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    List<Integer>[] graph = new ArrayList[n + 1];
    for (int i = 1; i <= n; i++) graph[i] = new ArrayList<>();

    for (int i = 0; i < n - 1; i++) {
      int u = sc.nextInt();
      int v = sc.nextInt();
      graph[u].add(v);
      graph[v].add(u);
    }

    int endpoint = bfs(graph, 1)[0];
    int diameter = bfs(graph, endpoint)[1];
    System.out.println(diameter);
  }
}

Python Solution

from collections import deque

n = int(input())
graph = [[] for _ in range(n + 1)]
for _ in range(n - 1):
    u, v = map(int, input().split())
    graph[u].append(v)
    graph[v].append(u)

def bfs(start):
    dist = [-1] * (n + 1)
    dist[start] = 0
    q = deque([start])
    while q:
        node = q.popleft()
        for nxt in graph[node]:
            if dist[nxt] == -1:
                dist[nxt] = dist[node] + 1
                q.append(nxt)
    farthest = max(range(1, n + 1), key=lambda x: dist[x])
    return farthest, dist[farthest]

endpoint, _ = bfs(1)
_, diameter = bfs(endpoint)
print(diameter)

Practice Challenge

Make the idea your own

Create a two-minute explanation of Tree Data Structure: define it, trace one example, name one edge case, and give the runtime of the main operation.

Tutoring Connection

Need help with this topic?

Bring a question or code attempt to a tutoring session.

Reserve a Free SessionText UsOpen Online Compiler