Diameter of Binary Tree in DSA Javascript - Time & Space Complexity

Time Complexity: Diameter of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the diameter of a binary tree changes as the tree grows.

Specifically, how does the number of steps grow when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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 1 + Math.max(left, right);
  }
  depth(root);
  return diameter;
}
    

This code finds the longest path between any two nodes in a binary tree by calculating depths recursively.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to visit each node once.
• How many times: Each node is visited exactly one time.

How Execution Grows With Input

As the number of nodes increases, the function visits each node once, so the total steps grow directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The steps grow in a straight line 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 function visits nodes multiple times, so it must be slower than linear."

[OK] Correct: Each node is visited only once during the depth calculation, so the total work is proportional to the number of nodes.

Interview Connect

Understanding this time complexity helps you explain how recursive tree algorithms scale, a common skill interviewers look for.

Self-Check

"What if we changed the code to calculate diameter by separately computing depths for each node without caching? How would the time complexity change?"

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