Diameter of Binary Tree in DSA Go

Mental Model

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

Analogy: Imagine a tree as a network of roads connecting cities; the diameter is the longest possible route you can take between two cities without repeating roads.

      1
     / \
    2   3
   / \
  4   5

Diameter could be path 4 -> 2 -> 1 -> 3 or 5 -> 2 -> 1 -> 3

Dry Run Walkthrough

Input: Binary tree: 1 with left child 2 and right child 3; 2 has children 4 and 5

Goal: Find the longest path between any two nodes in the tree

Step 1: Start at node 4, calculate height = 1 (leaf node)

Node 4: height=1, diameter=0

Why: Leaf nodes have height 1 and no diameter path inside

Step 2: Start at node 5, calculate height = 1 (leaf node)

Node 5: height=1, diameter=0

Why: Leaf nodes have height 1 and no diameter path inside

Step 3: At node 2, calculate height = max(1,1)+1=2; diameter through 2 = 1+1=2

Node 2: height=2, diameter=2 (path 4 -> 2 -> 5)

Why: Diameter through node 2 is sum of left and right heights

Step 4: At node 3, height = 1 (leaf), diameter=0

Node 3: height=1, diameter=0

Why: Leaf node height and diameter

Step 5: At root node 1, height = max(2,1)+1=3; diameter through 1 = 2+1=3

Node 1: height=3, diameter=3 (path 4 -> 2 -> 1 -> 3)

Why: Diameter through root is sum of left and right heights

Result:

Diameter = 3 (longest path: 4 -> 2 -> 1 -> 3)

Annotated Code

DSA Go

package main

import "fmt"

// TreeNode represents a node in the binary tree
type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

// diameterOfBinaryTree returns the diameter of the binary tree
func diameterOfBinaryTree(root *TreeNode) int {
    maxDiameter := 0

    // height calculates height of node and updates maxDiameter
    var height func(node *TreeNode) int
    height = func(node *TreeNode) int {
        if node == nil {
            return 0
        }
        leftHeight := height(node.Left) // height of left subtree
        rightHeight := height(node.Right) // height of right subtree

        // diameter through current node is sum of left and right heights
        if leftHeight+rightHeight > maxDiameter {
            maxDiameter = leftHeight + rightHeight
        }

        // height of current node is max of left/right heights plus 1
        return max(leftHeight, rightHeight) + 1
    }

    height(root)
    return maxDiameter
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

func main() {
    // Construct the tree:
    //      1
    //     / \
    //    2   3
    //   / \
    //  4   5
    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 := diameterOfBinaryTree(root)
    fmt.Println(diameter)
}

if node == nil { return 0 }

base case: empty node has height 0

leftHeight := height(node.Left)

recursively find left subtree height

rightHeight := height(node.Right)

recursively find right subtree height

if leftHeight+rightHeight > maxDiameter { maxDiameter = leftHeight + rightHeight }

update max diameter if path through current node is longer

return max(leftHeight, rightHeight) + 1

return height of current node for parent's calculation

Output

3

Complexity Analysis

Time: O(n) because each node is visited once during height calculation

Space: O(h) where h is tree height due to recursion stack

vs Alternative: Naive approach might compute height repeatedly causing O(n^2) time; this approach computes height and diameter in one pass

Edge Cases

Empty tree (root is nil)

Returns diameter 0 as there are no nodes

DSA Go

if node == nil { return 0 }

Single node tree

Diameter is 0 because no edges exist

DSA Go

if node == nil { return 0 }

Skewed tree (all nodes only on one side)

Diameter equals number of edges in the longest path (height - 1)

DSA Go

if leftHeight+rightHeight > maxDiameter { maxDiameter = leftHeight + rightHeight }

When to Use This Pattern

When asked for longest path between any two nodes in a tree, use diameter pattern by combining heights of left and right subtrees at each node.

Common Mistakes

Mistake: Calculating diameter as height of left subtree plus height of right subtree only at root node
Fix: Calculate diameter at every node during height calculation to find global maximum

Summary

Finds the longest path between any two nodes in a binary tree.

Use when you need the maximum distance between any two nodes, not just from root.

The diameter is the largest sum of left and right subtree heights at any node.

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