Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

Right View of a Binary Tree - Iterative Approach

Problem Statement

Given a binary tree, return the right view of the tree using an iterative level-order traversal approach. The right view of a binary tree includes the last node at each level when the tree is viewed from the right side.

Examples

Input Tree Right View Output Description
[1, 2, 3, 4, 5, null, 6]
[1, 3, 6] Standard case: Nodes visible from the right at each level
[1]
[1] Single-node tree: root is the only visible node
[] [] Empty tree: no nodes to view from any side
[1, 2, null, 3, null, null, null, 4]
[1, 2, 3, 4] Left-skewed tree: right view includes one node per level
[1, null, 2, null, null, null, 3]
[1, 2, 3] Right-skewed tree: all nodes are visible in the right view
[10, 20, 30, null, 25, null, 35]
[10, 30, 35] Balanced tree with both left and right children

Visualization Player

Solution

Understanding the Problem

The Right View of a Binary Tree refers to the set of nodes visible when the tree is viewed from the right side. At each level of the tree, we only see the rightmost node. Our task is to return a list of such nodes, ordered from top to bottom.

To solve this, we need to understand how trees are structured and how to explore them level by level. We will use a level-order traversal approach (like BFS), and for each level, we will record the last node encountered—this node represents what we see from the right side at that level.

Step-by-Step Solution with Example

Step 1: Understand the input format

The binary tree is typically represented as a level-order list like [1, 2, 3, null, 5, null, 4]. Each element corresponds to a node in the tree, and null indicates a missing child.

Step 2: Perform level-order traversal

We use a queue to explore the tree level by level. For each level, we loop through all nodes, and we keep track of the last node encountered in that level—this is the one visible from the right side.

Step 3: Apply the logic on the example

Let’s consider the tree represented by [1, 2, 3, null, 5, null, 4]. This corresponds to:

    1
   /   2   3
           5    4

Now we traverse level by level:

  • Level 1: Nodes = [1] → Rightmost = 1
  • Level 2: Nodes = [2, 3] → Rightmost = 3
  • Level 3: Nodes = [5, 4] → Rightmost = 4

So the final right view is [1, 3, 4].

Step 4: Store the result

As we detect the rightmost node at each level, we append it to a result list. After processing all levels, we return this list as our right view.

Edge Cases

Case 1: Tree is empty

If the binary tree has no nodes, then there is nothing to view. The right view is simply an empty list: [].

Case 2: Tree has only one node

When the tree contains just a single root node, that node is by default the right view. Example: [1] → Right view: [1].

Case 3: Tree is right skewed

If every node has only a right child, the entire tree is visible from the right. Example: [1, null, 2, null, 3] → Right view: [1, 2, 3].

Case 4: Tree is left skewed

Even though the nodes are on the left, each level has only one node, making it the rightmost by default. Example: [1, 2, null, 3] → Right view: [1, 2, 3].

Case 5: Tree has both left and right children

This is the general case where our logic must correctly select the last node from each level during traversal. The method we implemented ensures this behavior.

Finally

The key idea is to visit the tree level by level and pick the last node at each level. This last node is what you would see from the right. By using a queue and simple bookkeeping, we can ensure the solution works for skewed trees, balanced trees, and even empty trees.

This problem is a great introduction to BFS traversal and thinking about trees from different perspectives. Once you understand level-order traversal, the rest of the solution becomes very intuitive!

Algorithm Steps

  1. Given a binary tree, if the tree is empty, return an empty result.
  2. Initialize a queue and enqueue the root node.
  3. While the queue is not empty, determine the number of nodes at the current level (levelSize).
  4. For each node in the current level, dequeue a node.
  5. If the node is the last one in the level (i.e. for index i == levelSize - 1), record its value as part of the right view.
  6. Enqueue the left and then right child of the node (if they exist) for processing in the next level.
  7. Repeat until all levels are processed; the recorded values form the right view of the binary tree.

Code

C
C++
Python
Java
JS
Go
Rust
Kotlin
TS
#include <stdio.h>
#include <stdlib.h>

typedef struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
} TreeNode;

TreeNode* createNode(int val) {
    TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode));
    node->val = val;
    node->left = node->right = NULL;
    return node;
}

int* rightSideView(TreeNode* root, int* returnSize) {
    if (!root) {
        *returnSize = 0;
        return NULL;
    }
    int* result = malloc(100 * sizeof(int));
    TreeNode** queue = malloc(100 * sizeof(TreeNode*));
    int front = 0, rear = 0, resIndex = 0;
    queue[rear++] = root;
    while (front < rear) {
        int levelSize = rear - front;
        for (int i = 0; i < levelSize; i++) {
            TreeNode* node = queue[front++];
            if (i == levelSize - 1)
                result[resIndex++] = node->val;
            if (node->left) queue[rear++] = node->left;
            if (node->right) queue[rear++] = node->right;
        }
    }
    *returnSize = resIndex;
    free(queue);
    return result;
}

int main() {
    TreeNode* root = createNode(1);
    root->left = createNode(2);
    root->right = createNode(3);
    root->left->right = createNode(5);
    root->right->right = createNode(4);
    int size;
    int* view = rightSideView(root, &size);
    printf("Right view: ");
    for (int i = 0; i < size; i++) printf("%d ", view[i]);
    printf("\n");
    free(view);
    return 0;
}

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