Diameter of Binary Tree in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to calculate the height of a binary tree node.

DSA Javascript

function height(node) {
  if (node === null) return 0;
  return 1 + Math.max(height(node.left), height(node.[1]));
}

Drag options to blanks, or click blank then click option'

Aparent

Broot

Cright

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' instead of 'right' causes undefined errors.

Using 'child' or 'root' are not valid properties of a node.

The height function calculates the height of a node by recursively checking the left and right children. The correct property to access the right child is 'right'.

2fill in blank

medium

Complete the code to update the diameter during traversal.

DSA Javascript

function diameterOfBinaryTree(root) {
  let diameter = 0;
  function dfs(node) {
    if (node === null) return 0;
    let leftHeight = dfs(node.left);
    let rightHeight = dfs(node.right);
    diameter = Math.max(diameter, leftHeight + rightHeight);
    return 1 + Math.max(leftHeight, [1]);
  }
  dfs(root);
  return diameter;
}

Drag options to blanks, or click blank then click option'

Anode

Bdiameter

Croot

DrightHeight

Attempts:

3 left

💡 Hint

Common Mistakes

Returning diameter instead of rightHeight causes wrong height calculation.

Returning node or root is incorrect as they are objects, not heights.

The function returns the height of the current node, which is 1 plus the maximum height of its left and right children. So, it should return 1 + Math.max(leftHeight, rightHeight).

3fill in blank

hard

Fix the error in the recursive function call to avoid infinite recursion.

DSA Javascript

function dfs(node) {
  if (node === null) return 0;
  let leftHeight = dfs(node.left);
  let rightHeight = dfs(node.[1]);
  return 1 + Math.max(leftHeight, rightHeight);
}

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cparent

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'left' twice causes missing right subtree traversal.

Using 'parent' causes infinite recursion or errors.

The recursive call should be on the right child of the node, which is accessed by 'right'. Using 'parent' or 'node' would cause errors or infinite recursion.

4fill in blank

hard

Fill both blanks to correctly update the diameter and return the height.

DSA Javascript

function dfs(node) {
  if (node === null) return 0;
  let leftHeight = dfs(node.left);
  let rightHeight = dfs(node.right);
  diameter = Math.max(diameter, leftHeight [1] rightHeight);
  return 1 [2] Math.max(leftHeight, rightHeight);
}

Drag options to blanks, or click blank then click option'

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-' or '*' causes wrong diameter or height calculations.

Using '/' causes runtime errors or wrong results.

The diameter is updated by adding leftHeight and rightHeight. The height is 1 plus the maximum of leftHeight and rightHeight, so both blanks use '+'.

5fill in blank

hard

Fill all three blanks to complete the diameter calculation using a helper function.

DSA Javascript

function diameterOfBinaryTree(root) {
  let diameter = 0;
  function dfs(node) {
    if (node === null) return 0;
    let leftHeight = dfs(node.left);
    let rightHeight = dfs(node.right);
    diameter = Math.max(diameter, leftHeight [1] rightHeight);
    return 1 + Math.max(leftHeight, [2]);
  }
  dfs([3]);
  return diameter;
}

Drag options to blanks, or click blank then click option'

A+

BrightHeight

Croot

DleftHeight

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong operators like '-' or '*' for diameter update.

Returning wrong variable instead of rightHeight.

Calling dfs on leftHeight or rightHeight instead of root.

The diameter is updated by adding leftHeight and rightHeight (use '+'). The height returned is 1 plus the max of leftHeight and rightHeight. The dfs function is called on the root node.

Practice

1. What does the diameter of a binary tree represent?

easy

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

B. The height of the tree

C. The number of leaf nodes

D. The total number of nodes in the tree

Solution

1. Step 1: Understand the diameter concept

   The diameter is the longest path between any two nodes, which may or may not pass through the root.

2. Step 2: Differentiate from height and node count

   Height is the longest path from root to leaf, and node count is total nodes, both different from diameter.

3. Final Answer:

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

4. Quick Check:

   Diameter = Longest path between nodes [OK]

Hint: Diameter is longest path, not just height or node count [OK]

Common Mistakes:

• Confusing diameter with tree height
• Thinking diameter counts nodes instead of edges
• Assuming diameter must pass through root

2. Which of the following JavaScript function signatures is correct for computing the diameter of a binary tree node?

easy

A. function diameterTree() {}

B. function diameter(root, tree) {}

C. function binaryTreeDiameter(node, root) {}

D. function diameterOfBinaryTree(root) {}

Solution

1. Step 1: Check common parameter naming

   The function usually takes the root node as input, so a single parameter named root is standard.

2. Step 2: Validate function signature correctness

   function diameterOfBinaryTree(root) {} uses a clear, single parameter 'root' and a descriptive function name, matching common practice.

3. Final Answer:

   function diameterOfBinaryTree(root) {} -> Option D

4. Quick Check:

   Correct function signature uses root parameter [OK]

Hint: Function usually takes root node as single parameter [OK]

Common Mistakes:

• Using multiple unrelated parameters
• Incorrect or unclear function names
• Missing parameters for the tree root

3. Given the following binary tree structure and code, what is the diameter of the 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,
    new TreeNode(4),
    new TreeNode(5)
  ),
  new TreeNode(3)
);

function diameterOfBinaryTree(root) {
  let diameter = 0;
  function depth(node) {
    if (!node) return 0;
    const left = depth(node.left);
    const right = depth(node.right);
    diameter = Math.max(diameter, left + right);
    return Math.max(left, right) + 1;
  }
  depth(root);
  return diameter;
}

console.log(diameterOfBinaryTree(root));

medium

A. 4

B. 3

C. 2

D. 5

Solution

1. Step 1: Analyze the tree structure

   The tree has root 1, left subtree with nodes 2,4,5 and right subtree with node 3. The longest path is from node 4 to node 3 through 2 and 1 (4->2->1->3), which has length 3 edges.

2. Step 2: Trace the code logic

   The code calculates diameter as max sum of left and right depths at each node. At node 2, left and right depths are 1 each, so diameter updates to 2. At root, left depth is 2, right depth is 1, so diameter updates to 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Longest path edges = 3 [OK]

Hint: Diameter = max sum of left and right subtree depths [OK]

Common Mistakes:

• Counting nodes instead of edges for diameter
• Ignoring paths not passing through root
• Misreading tree structure

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

function diameterOfBinaryTree(root) {
  let diameter = 0;
  function depth(node) {
    if (!node) return 0;
    const left = depth(node.left);
    const right = depth(node.right);
    diameter = Math.max(diameter, left * right);
    return Math.max(left, right) + 1;
  }
  depth(root);
  return diameter;
}

medium

A. Using multiplication instead of addition for diameter update

B. Returning max depth instead of sum of depths

C. Not handling null nodes correctly

D. Missing base case for leaf nodes

Solution

1. Step 1: Identify diameter update logic

   The code updates diameter using left * right, which multiplies depths instead of adding them.

2. Step 2: Correct diameter calculation

   Diameter should be updated with left + right, representing the longest path through the node.

3. Final Answer:

   Using multiplication instead of addition for diameter update -> Option A

4. Quick Check:

   Diameter update uses addition, not multiplication [OK]

Hint: Diameter update = left depth + right depth, not multiplication [OK]

Common Mistakes:

• Multiplying depths instead of adding
• Confusing depth return with diameter update
• Ignoring base case for null nodes

5. You want to find the diameter of a binary tree but also return the path nodes that form this diameter. Which approach correctly extends the diameter calculation to return the path?

hard

A. Use breadth-first search to find the longest path from root to any leaf

B. Only compute the height of the tree and return the nodes at the deepest level

C. Track left and right subtree depths and their paths; update diameter and store combined path when a longer diameter is found

D. Count total nodes and return all leaf nodes as the diameter path

Solution

1. Step 1: Understand diameter path requirements

   Diameter path is the longest path between any two nodes, so we must track paths from left and right subtrees.

2. Step 2: Extend depth function to track paths

   Along with depths, store paths for left and right subtrees. When diameter updates, combine these paths to form the diameter path.

3. Step 3: Reject incorrect approaches

   Height alone or BFS from root won't find longest path between any two nodes. Counting leaf nodes is unrelated.

4. Final Answer:

   Track left and right subtree depths and their paths; update diameter and store combined path when a longer diameter is found -> Option C

5. Quick Check:

   Track depths and paths to get diameter path [OK]

Hint: Store paths with depths; update when diameter increases [OK]

Common Mistakes:

• Only tracking height or BFS from root
• Ignoring path storage during recursion
• Assuming diameter path is always root to leaf