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DSA Javascriptprogramming~5 mins

Right Side View of Binary Tree in DSA Javascript - Cheat Sheet & Quick Revision

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Recall & Review
beginner
What is the Right Side View of a binary tree?
It is the list of nodes visible when looking at the tree from the right side, showing the rightmost node at each level.
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beginner
Which traversal method is commonly used to get the right side view of a binary tree?
Level order traversal (Breadth-First Search) is used, visiting nodes level by level and recording the last node at each level.
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beginner
In the right side view algorithm, why do we record the last node at each level?
Because the last node visited at each level is the rightmost node visible from the right side.
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intermediate
What data structure helps to implement the right side view using level order traversal?
A queue is used to keep track of nodes at each level during the traversal.
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intermediate
How does the recursive approach for right side view differ from the iterative approach?
The recursive approach uses depth-first search prioritizing the right child first, recording the first node visited at each depth, while iterative uses BFS and records the last node at each level.
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What does the right side view of a binary tree show?
AThe rightmost nodes at each level
BThe root node only
CAll leaf nodes
DThe leftmost nodes at each level
Which data structure is typically used in the iterative method to find the right side view?
AStack
BSet
CQueue
DMap
In a recursive right side view approach, which child is visited first?
ALeft child
BRight child
CBoth children simultaneously
DNo children
What is the time complexity of the right side view algorithm using BFS?
AO(n)
BO(log n)
CO(n^2)
DO(1)
If a binary tree has only left children, what will the right side view look like?
AOnly leaf nodes
BEmpty list
COnly the root node
DAll nodes in the tree
Explain how to find the right side view of a binary tree using level order traversal.
Think about visiting nodes level by level and picking the rightmost node.
You got /4 concepts.
    Describe the difference between iterative and recursive approaches to get the right side view of a binary tree.
    Consider traversal order and how nodes are recorded.
    You got /4 concepts.

      Practice

      (1/5)
      1. What does the Right Side View of a binary tree represent?
      easy
      A. The nodes at the deepest level of the tree
      B. The nodes visible when the tree is viewed from the right side
      C. All leaf nodes of the tree
      D. The nodes visible when the tree is viewed from the left side

      Solution

      1. Step 1: Understand the view perspective

        The right side view shows nodes visible when looking at the tree from the right side.
      2. Step 2: Identify visible nodes

        Only the rightmost node at each level is visible from the right side.
      3. Final Answer:

        The nodes visible when the tree is viewed from the right side -> Option B
      4. Quick Check:

        Right side view = rightmost nodes at each level [OK]
      Hint: Right side view shows rightmost nodes at each tree level [OK]
      Common Mistakes:
      • Confusing right side view with left side view
      • Thinking it shows all leaf nodes
      • Assuming it shows deepest nodes only
      2. Which of the following JavaScript function signatures correctly represents a function to get the right side view of a binary tree?
      easy
      A. function rightSideView() { }
      B. function rightSideView(root, result) { }
      C. function rightSideView(root) { }
      D. function rightSideView(tree, level) { }

      Solution

      1. Step 1: Check typical function signature

        The function usually takes the root node as input to start traversal.
      2. Step 2: Validate options

        function rightSideView(root) { } correctly takes only the root node, which is standard for this problem.
      3. Final Answer:

        function rightSideView(root) { } -> Option C
      4. Quick Check:

        Function takes root node only [OK]
      Hint: Function usually takes root node as single parameter [OK]
      Common Mistakes:
      • Adding unnecessary parameters
      • Omitting root parameter
      • Using unclear parameter names
      3. What is the output of the following code snippet for the given binary tree?
      class TreeNode {
        constructor(val, left = null, right = null) {
          this.val = val;
          this.left = left;
          this.right = right;
        }
      }
      
      const root = new TreeNode(1,
        new TreeNode(2, null, new TreeNode(5)),
        new TreeNode(3, null, new TreeNode(4))
      );
      
      function rightSideView(root) {
        const result = [];
        const queue = [root];
        while (queue.length) {
          const size = queue.length;
          for (let i = 0; i < size; i++) {
            const node = queue.shift();
            if (i === size - 1) result.push(node.val);
            if (node.left) queue.push(node.left);
            if (node.right) queue.push(node.right);
          }
        }
        return result;
      }
      
      console.log(rightSideView(root));
      medium
      A. [1, 3, 4]
      B. [1, 2, 5]
      C. [1, 3, 5]
      D. [1, 2, 4]

      Solution

      1. Step 1: Traverse level by level and identify rightmost nodes

        Level 1: Node 1 (only node), Level 2: Nodes 2 and 3 (rightmost is 3), Level 3: Nodes 5 and 4 (rightmost is 4).
      2. Step 2: Collect rightmost nodes in order

        Result array is [1, 3, 4].
      3. Final Answer:

        [1, 3, 4] -> Option A
      4. Quick Check:

        Rightmost nodes per level = [1, 3, 4] [OK]
      Hint: Pick last node at each level during BFS [OK]
      Common Mistakes:
      • Choosing leftmost nodes instead of rightmost
      • Mixing node values from different levels
      • Forgetting to check all nodes at a level
      4. Identify the bug in the following code that attempts to find the right side view of a binary tree:
      function rightSideView(root) {
        const result = [];
        const queue = [root];
        while (queue.length) {
          const size = queue.length;
          for (let i = 0; i < size; i++) {
            const node = queue.shift();
            if (i === 0) result.push(node.val);
            if (node.left) queue.push(node.left);
            if (node.right) queue.push(node.right);
          }
        }
        return result;
      }
      medium
      A. It pushes the leftmost node instead of the rightmost node at each level
      B. It does not check if root is null
      C. It uses shift() which is inefficient
      D. It pushes both left and right children in wrong order

      Solution

      1. Step 1: Analyze the condition for pushing node values

        The code pushes node.val when i === 0, which is the first node at each level (leftmost).
      2. Step 2: Understand right side view requirement

        Right side view requires the rightmost node at each level, which is when i === size - 1.
      3. Final Answer:

        It pushes the leftmost node instead of the rightmost node at each level -> Option A
      4. Quick Check:

        Right side view needs last node per level, not first [OK]
      Hint: Push node when i equals size-1 for right side view [OK]
      Common Mistakes:
      • Using i === 0 instead of i === size - 1
      • Ignoring null root check
      • Confusing order of child insertion
      5. Given a binary tree where some nodes have only left children, how would the right side view look? Example tree:
          1
         /
        2
       / 
      3
      
      What is the right side view output for this tree?
      hard
      A. [3]
      B. [1]
      C. [1, 3]
      D. [1, 2, 3]

      Solution

      1. Step 1: Visualize the tree and right side view

        Since all nodes are on the left, the right side view shows these nodes as they are the only visible ones.
      2. Step 2: List nodes level-wise

        Level 1: 1, Level 2: 2, Level 3: 3. All are visible from right side as no right nodes block view.
      3. Final Answer:

        [1, 2, 3] -> Option D
      4. Quick Check:

        Right side view includes all nodes when only left children exist [OK]
      Hint: If no right child, left child is visible from right side [OK]
      Common Mistakes:
      • Assuming right side view is empty if no right children
      • Skipping left children in right side view
      • Confusing right side view with right subtree nodes only
      Recommended

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