Diameter of Binary Tree in DSA Go - Execution Trace

Concept Flow - Diameter of Binary Tree

Start at root node

↓

Compute left subtree height

↓

Compute right subtree height

↓

Calculate diameter at this node = left height + right height

↓

Update max diameter if current is larger

↓

Return height = max(left height, right height) + 1

↓

Repeat for left and right child nodes recursively

↓

Final max diameter is the answer

We recursively find the height of left and right subtrees at each node, update the maximum diameter found, and return the height to parent nodes.

Execution Sample

DSA Go

func diameterOfBinaryTree(root *TreeNode) int {
  maxDiameter := 0
  var height func(node *TreeNode) int
  height = func(node *TreeNode) int {
    if node == nil { return 0 }
    left := height(node.Left)
    right := height(node.Right)
    if left + right > maxDiameter { maxDiameter = left + right }
    if left > right {
      return left + 1
    }
    return right + 1
  }
  height(root)
  return maxDiameter
}

This code calculates the diameter of a binary tree by recursively computing subtree heights and tracking the maximum sum of left and right heights.

Execution Table

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

💡 Traversal complete, max diameter found is 4 (longest path between two nodes).

Variable Tracker

Variable	Start	After Step 3	After Step 6	After Step 7	After Step 11	After Step 14	After Step 16	Final
maxDiameter	0	0	0	2	2	2	4	4
left (height)	?	1	1	1	1	1	2	2
right (height)	?	?	1	1	1	1	2	2
return height	?	1	1	2	1	2	3	3

Key Moments - 3 Insights

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

Because the diameter is the longest path through that node, which is the sum of the longest path down the left subtree and the longest path down the right subtree. See execution_table rows 7, 14, and 16 where diameter at node is left height + right height.

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

Height is the longest path from the node down to a leaf. We take the larger height of left or right subtree and add 1 for the current node. This is shown in execution_table rows 4, 6, 8, 11, 13, 15, and 17.

How does maxDiameter get updated during recursion?

At each node, we check if left height + right height is greater than current maxDiameter and update it. This happens in rows 7, 14, and 16 where maxDiameter changes from 0 to 2 and then to 4.

Visual Quiz - 3 Questions

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

A 0

B 1

C 2

D 3

💡 Hint

Check the 'Diameter at Node' column in row 7 of execution_table.

At which step does maxDiameter first update from 0 to 2?

A Step 7

B Step 3

C Step 14

D Step 16

💡 Hint

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

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

A It would be zero

B It would be left height + right height (which is zero)

C It would be equal to left height

D It would be one

💡 Hint

Recall diameter at node is left height + right height; if right height is zero, diameter equals left height + 0.

Concept Snapshot

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

Full Transcript

The diameter of a binary tree is the longest path between any two nodes. We find it by recursively calculating the height of left and right subtrees at each node. The diameter at a node is the sum of left and right subtree heights. We keep track of the maximum diameter found during traversal. The height returned to parent nodes is the maximum of left and right heights plus one for the current node. This process continues until all nodes are visited, and the maximum diameter found is returned as the result.

Practice

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

easy

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

B. The number of nodes in the tree

C. The height of the tree

D. The number of leaf nodes in the tree

Solution

1. Step 1: Understand the diameter definition

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

2. Step 2: Differentiate from other tree properties

   Height is the longest path from root to leaf, number of nodes counts all nodes, leaves count only leaf nodes.

3. Final Answer:

   The length of 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 height or node count [OK]

Common Mistakes:

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

2. Which Go function signature correctly represents a helper function to compute the height of a binary tree node for diameter calculation?

easy

A. func height(node int) int

B. func height(node *TreeNode) int

C. func height(node *TreeNode) bool

D. func height(node TreeNode) string

Solution

1. Step 1: Identify correct parameter type

   The function should accept a pointer to TreeNode to handle nil nodes and avoid copying.

2. Step 2: Identify correct return type

   Height is an integer value representing depth, so return type must be int.

3. Final Answer:

   func height(node *TreeNode) int -> Option B

4. Quick Check:

   Pointer input and int output for height [OK]

Hint: Height function uses pointer receiver and returns int [OK]

Common Mistakes:

• Using value receiver instead of pointer
• Returning wrong type like string or bool
• Passing int instead of TreeNode pointer

3. Given the following Go code snippet, what is the diameter of the binary tree?

type TreeNode struct {
    Val int
    Left *TreeNode
    Right *TreeNode
}

root := &TreeNode{Val: 1}
root.Left = &TreeNode{Val: 2}
root.Right = &TreeNode{Val: 3}
root.Left.Left = &TreeNode{Val: 4}
root.Left.Right = &TreeNode{Val: 5}

// Diameter function uses height helper internally
func diameterOfBinaryTree(root *TreeNode) int {
    maxDiameter := 0
    var height func(node *TreeNode) int
    height = func(node *TreeNode) int {
        if node == nil {
            return 0
        }
        leftHeight := height(node.Left)
        rightHeight := height(node.Right)
        if leftHeight+rightHeight > maxDiameter {
            maxDiameter = leftHeight + rightHeight
        }
        if leftHeight > rightHeight {
            return leftHeight + 1
        }
        return rightHeight + 1
    }
    height(root)
    return maxDiameter
}

fmt.Println(diameterOfBinaryTree(root))

medium

A. 4

B. 2

C. 3

D. 5

Solution

1. Step 1: Calculate heights of subtrees

   Left subtree of root has height 2 (nodes 2->4 or 2->5), right subtree height is 1 (node 3).

2. Step 2: Calculate diameter as sum of left and right heights

   Diameter = leftHeight + rightHeight = 2 + 1 = 3 edges, but diameter counts edges, so longest path is 3 edges which means 4 nodes.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Longest path length = 3 edges [OK]

Hint: Diameter = sum of left and right subtree heights at max node [OK]

Common Mistakes:

• Counting nodes instead of edges for diameter
• Ignoring one subtree height
• Confusing height with diameter

4. Identify the bug in this Go code snippet for computing the diameter of a binary tree:

func diameterOfBinaryTree(root *TreeNode) int {
    maxDiameter := 0
    var height func(node *TreeNode) int
    height = func(node *TreeNode) int {
        if node == nil {
            return 0
        }
        leftHeight := height(node.Left)
        rightHeight := height(node.Right)
        maxDiameter = leftHeight + rightHeight
        if leftHeight > rightHeight {
            return leftHeight + 1
        }
        return rightHeight + 1
    }
    height(root)
    return maxDiameter
}

medium

A. Function does not call height recursively

B. height function returns wrong type

C. Missing base case for nil node

D. maxDiameter is overwritten without checking max value

Solution

1. Step 1: Analyze maxDiameter update

   maxDiameter is assigned leftHeight + rightHeight every time, overwriting previous max without comparison.

2. Step 2: Correct maxDiameter update

   Should update maxDiameter only if leftHeight + rightHeight is greater than current maxDiameter.

3. Final Answer:

   maxDiameter is overwritten without checking max value -> Option D

4. Quick Check:

   Use max check before assignment [OK]

Hint: Update maxDiameter only if new value is greater [OK]

Common Mistakes:

• Overwriting maxDiameter without max check
• Forgetting base case for nil node
• Not calling height recursively

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

hard

A. Modify height function to return both height and path, update max path when diameter increases

B. Calculate diameter first, then do a separate BFS to find path

C. Store all paths in a global slice and pick the longest after traversal

D. Use only height function and return diameter length without path

Solution

1. Step 1: Extend height function to track paths

   Modify the recursive height function to return both height and the path nodes from current node to leaf.

2. Step 2: Update max diameter and store path

   When leftHeight + rightHeight exceeds maxDiameter, update maxDiameter and store concatenated path from left and right subtrees through current node.

3. Step 3: Return stored path after traversal

   After full traversal, return the stored path representing the diameter.

4. Final Answer:

   Modify height function to return both height and path, update max path when diameter increases -> Option A

5. Quick Check:

   Track height and path together for diameter path [OK]

Hint: Return height and path from helper to track diameter path [OK]

Common Mistakes:

• Trying to find path after diameter without tracking
• Storing all paths wastes memory
• Ignoring path tracking in recursion