Diameter of Binary Tree in DSA Javascript - Execution Trace

Concept Flow - Diameter of Binary Tree

Start at root node

↓

Calculate left subtree height

↓

Calculate right subtree height

↓

Calculate diameter through current node = left height + right height

↓

Update max diameter if current diameter is larger

↓

Return height of current node = max(left height, right height) + 1

↓

Repeat for left and right child nodes recursively

↓

End when all nodes visited

↓

Final max diameter is result

We start at the root and recursively find the height of left and right subtrees. At each node, we calculate the diameter passing through it and update the maximum diameter found. We return the height to parent nodes to continue the process.

Execution Sample

DSA Javascript

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

This code calculates the diameter of a binary tree by recursively computing subtree heights and updating the maximum diameter found.

Execution Table

Step	Operation	Node Visited	Left Height	Right Height	Diameter Through Node	Max Diameter	Return Height	Tree State
1	Visit root	1	?	?	?	0	?	1
2	Visit left child	2	?	?	?	0	?	1 -> 2
3	Visit left-left child	4	0	0	0	0	1	1 -> 2 -> 4
4	Return from node 4	4	0	0	0	0	1	1 -> 2 -> 4
5	Visit left-right child	5	0	0	0	0	1	1 -> 2 -> 4,5
6	Return from node 5	5	0	0	0	0	1	1 -> 2 -> 4,5
7	Calculate node 2 diameter	2	1	1	2	2	2	1 -> 2 -> 4,5
8	Return from node 2	2	1	1	2	2	2	1 -> 2 -> 4,5
9	Visit right child	3	?	?	?	2	?	1 -> 2,3 -> 4,5
10	Visit right-left child	6	0	0	0	2	1	1 -> 2,3 -> 4,5,6
11	Return from node 6	6	0	0	0	2	1	1 -> 2,3 -> 4,5,6
12	Visit right-right child	7	0	0	0	2	1	1 -> 2,3 -> 4,5,6,7
13	Return from node 7	7	0	0	0	2	1	1 -> 2,3 -> 4,5,6,7
14	Calculate node 3 diameter	3	1	1	2	2	2	1 -> 2,3 -> 4,5,6,7
15	Return from node 3	3	1	1	2	2	2	1 -> 2,3 -> 4,5,6,7
16	Calculate root node diameter	1	2	2	4	4	3	1 -> 2,3 -> 4,5,6,7
17	Return from root	1	2	2	4	4	3	1 -> 2,3 -> 4,5,6,7
18	End	-	-	-	-	4	-	Final diameter = 4

💡 All nodes visited, maxDiameter updated to 4, recursion ends.

Variable Tracker

Variable	Start	After Step 3	After Step 6	After Step 7	After Step 11	After Step 13	After Step 14	After Step 16	Final
maxDiameter	0	0	0	2	2	2	2	4	4
height(node 4)	?	1	1	1	1	1	1	1	1
height(node 5)	?	?	1	1	1	1	1	1	1
height(node 2)	?	?	?	2	2	2	2	2	2
height(node 6)	?	?	?	?	1	1	1	1	1
height(node 7)	?	?	?	?	?	1	1	1	1
height(node 3)	?	?	?	?	?	?	2	2	2
height(node 1)	?	?	?	?	?	?	?	3	3

Key Moments - 3 Insights

Why do we add left height and right height to get diameter at a node?

Because diameter is the longest path between two nodes, which can pass through the current node connecting left and right subtrees. This is shown in execution_table rows 7, 14, and 16 where diameter through node is left height + right height.

Why do we return max(left height, right height) + 1 as height?

Height is the longest path from the node down to a leaf. We add 1 for the current node itself. This is tracked in variable_tracker for each node's height after visiting children.

Why does maxDiameter update only when left + right is larger?

Because diameter is the longest path found so far. We update maxDiameter only if the current node's diameter (left + right) is greater, as seen in execution_table steps 7, 14, and 16.

Visual Quiz - 3 Questions

Look at the execution_table at step 7, what is the diameter through node 2?

A 1

B 2

C 3

D 0

💡 Hint

Check the 'Diameter Through Node' column at step 7 in execution_table.

At which step does maxDiameter first update to 2?

A Step 7

B Step 3

C Step 6

D Step 14

💡 Hint

Look at the 'Max Diameter' column in execution_table and find when it changes from 0 to 2.

If the tree had no right subtree, how would maxDiameter change at root?

A It would be left subtree height + 1

B It would be equal to left subtree height

C It would be left subtree height

D It would be 0

💡 Hint

Diameter through node is left height + right height. If right height is 0, diameter equals left height.

Concept Snapshot

Diameter of Binary Tree:
- Diameter = longest path between any two nodes
- At each node, diameter = left subtree height + right subtree height
- Use recursion to get heights and update max diameter
- Return height = max(left, right) + 1
- Final max diameter is the answer

Full Transcript

The diameter of a binary tree is the longest path between any two nodes. We find it by visiting each node and calculating the height of its left and right subtrees. The diameter through a node is the sum of these heights. We keep track of the maximum diameter found so far. The height returned to parent nodes is the maximum of left and right heights plus one for the current node. This process repeats recursively until all nodes are visited. The final maximum diameter is returned as the result.

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